An Electrical Library* 

By PROF. T. O'CONOR SLOANE. 



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Arithmetic of Electricity. 

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Standard Electrical Dictionary. 

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NORMAN W. HENLEY & CO., Publishers, 

132 Nassau Street, New York. 



Arithmetic of Electricity 



BEING 
A PRACTICAL TREATISE} ON ELECTRICAL CALCULATIONS 
OF ALL KINDS REDUCED TO A SERIES OF RULES, 
ALL OF THE SIMPLEST FORMS, AND INVOLVING 
ONLY ORDINARY ARITHMETIC, EACH RULE 
ILLUSTRATED BY ONE OR MORE PRAC- 
TICAL PROBLEMS, WITH DETAILED 
SOLUTION OF EACH. FOL- 
LOWED BY AN EXTENSIVE \ * i » • \ 
SERIES OF TABLES ' 



BY 

T. O'CONOR SLOANE, A.M., E.M., Ph.D. 

AUTHOR OF 

Standard Electrical Dictionary, Electricity Simplified, 
Electric Toy Making, etc. 



Illustrated 
Sixteenth Edition, Revised and Enlarged 



NEW YORK 

NORMAN W. HENLEY & CO. 

132 Nassau Street 
1903 



THE LIBRARY OF 
CONGRESS, 

Two Copies Received 

JUt 30 1903 

v Copyright Entry 
CUS& <t XX&No, 
COPY 8. 



52 









Copyrighted, 1891, 
NORMAN W. HENLEY & CO. 



Copyrighted, 1903, 



NORMAN W. HENLEY & CO. 



PREFACE. 



The solution of a problem by arithmetic, although 
in some cases more laborious than the algebraic method, 
gives the better comprehension of the subject. Arith- 
metic is analysis and bears the same relation to algebra 
that plane geometry does to analytical geometry. Its 
power is comparatively limited, but it is exceedingly 
instructive in its treatment of questions to which it 
applies. 

In the following work the problems of electrical en- 
gineering and practical operations are investigated on 
an arithmetical basis. It is believed that such treatment 
gives the work actual value in the analytical sense, as 
it necessitates an explanation of each problem, while 
the adaptability of arithmetic to readers who do not 
care to use algebra will make this volume more widely 
available. 

In electricity there is much debatable ground, which 
has been as far as possible avoided. Some points seem 
quite outside of the scope of this book, such as the intro- 
duction of the time-constant in battery calculations. 
Again the variation in constants as determined by dif- 
ferent authorities made a selection embarrassing. It Is 
believed that some success has been attained in over- 
coming or compromising difficulties such as those sug- 
gested. 



iv PREFACE. 

Enough tables have ^een introduced to fill the limits 
of the subject as here treated. 

The full development of electrical laws involves the 
higher mathematics. One who would keep up with the 
progress of the day in theory has a severe course of 
study before him. In practical work it is believed that 
such a volume as the Arithmetic of Electricity will 
always have a place. We hope that it will be favorably 
received by our readers and that their indulgence will 
give it a more extended field of usefulness than it can 
pretend to deserve. 



PREFACE TO SIXTEENTH EDITION. 

The steady progress of electrical science in conjunc- 
tion with a continued demand for this work have made 
advisable a revision and extension of this book. 

The author feels that in the matter which has been 
added much more could have been said on the subjects 
treated of, but, since a full exposition of each theme 
would alone fill a volume, it is hoped that the practical 
value of the rules, etc., will atone for the brevity of the 
text. 

In the preparation of this edition the author would 
express his indebtedness to A. A. Atkinson's excellent 
work on Electrical and Magnetic Calculations and also 
to the instruction papers of the Electrical Engineering 
Course of the International Correspondence School of 
Scranton, Pa. He would also express his thanks to 
Henry V. A. Parsell, for his valued advice and assistance 
in the preparation of the manuscript. 

The Authob. 

June, 1903. 



CONTENTS. 



CHAPTER I. 

INTRODUCTORY. 

Space, Time, Force, Resistance, Work, Energy, Mass 
and Weight.— The Fundamental Units of Dimension, 
and Derived Units, Geometrical, Mechanical, and Elec- 
trical.— C. G. S. and Practical Electrical Units— Nomen- 
clature.— Examples from Actual Practice 9 

CHAPTER II. 

ohm's law. 

General Statement. — Six Rules Derived by Transpos- 
ition from the Law. — Single Conductor Closed Circuits. 
— Batteries in Opposition. — Portions of Circuits.— Di- 
vided Circuits, with Calculation of Currents Passed by 
Each Branch, and of their Combined Resistance 13 

CHAPTER III. 

RESISTANCE AND CONDUCTANCE. 

Resistance of Different Conductors of the same Mater- 
ial. — Relations of Wires of Equal Resistance. — Ratio of 
Resistance of Two Conductors. — Specific Resistance. — 
Universal Rule for Resistances. — Resistance of Wires 
Referred to Weight. — Conductance. — Ohm's Law ex- 
pressed in Conductance 26 



vi CONTENTS. 

CHAPTER IV. 

POTENTIAL DIFFERENCE. 

Drop of Potential in Leads, and Size of Same for Mul- 
tiple Arc Connections. — Diminishing Size of Leads 
Progressively 38 

CHAPTER V. 

CIRCULAR MILS. 

The Mil.— The Circular Mil as a Unit of Area.— Cir- 
cular Mil Rules for Resistance and Size of Leads 43 



CHAPTER VI. 

SPECIAL SYSTEMS. 

Three Wire System.— Rules for Calculating Leads in 
Same.— Alternating Current System. — Ratio of Conver- 
sion.— Size of Primary Wire.— Converter Winding 46 



CHAPTER VII. 

WORK AND ENERGY. 

Energy and Heating Effect of the Current.— Differ- 
ent Rules Based on Joule's Law. — The Joule or Gram- 
Calorie. — Quantity of Heat Developed in an Active Cir- 
cuit in a Unit of Time. — Watts and Amperes in Rela- 
tion to Time.— Specific Heat. — Heating of Wire by a 
Current.— Safety Fuses.— Work of a Current. — Elec- 
trical Horse-Power. — Duty and Efficiency of Electrical 
Generators 50 



CONTENTS. vli 

CHAPTER VIII. 

BATTERIES. 

Arrangement of Battery Cells. — General Calcula- 
tions of Current. — Rules for Arrangement of Cells 
in a Battery. — Battery Calculations for Specified 
Electromotive Force and Current. — Efficiency of 
Batteries. — Chemistry of Batteries. — Calculation 
of Voltage. — Work of Batteries. — Efficiency of Bat- 
teries, to Calculate. — Chemicals Consumed in a 
Battery. — Decomposition of Compounds by a Bat- 
tery. — Electroplating 6G 

CHAPTER IX. 

ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 

The Magnetic Field and Lines of Force. — Per- 
meance and Reluctance. — Magnetizing Force and 
the Magnetic Circuit. — General Rules for Electro- 
Magnets and Ampere-Turns for Given Magnetic 
Flux. — Magnetic Circuit Calculations. — Leakage of 
Lines of Force. — Example of Calculation of a Mag- 
netic Circuit. — Dynamo Armatures. — Voltage and 
Capacity of Armatures.— Drum Type Closed Cir- 
cuit Armatures. — Field Magnets of Dynamos. — The 
Kapp Line 82 

CHAPTER X. 

ELECTRIC RAILWAYS. 

Sizes of Feeders. — Power to Move Cars 109 



viii CONTENTS. 

CHAPTER XL 

ALTERNATING CURRENTS. 

Self-induction 115 

CHAPTER XII. 

CONDENSERS. 

Page 121 

CHAPTER XIII. 



DEMONSTRATION OF RULES. 

Some of the Principal Rules in the Work Demon- 
strated 127 



CHAPTER XIV. 

NOTATION IN POWERS OF TEN. 

The Four Fundamental Operations of Addition, 
Subtraction, Multiplication, and Division in Powers 
of Ten 136 

TABLES. 

A Collection of Tables Needed for the Operations 
and Problems Given in the Work 139 



ARITHMETIC OF ELECTRICITY. 



CHAPTER I. 

INTRODUCTORY. 



Space is the lineal distance from one point to 
another. 

Time is the measure of duration. 

Force is any cause of change of motion of matter. 
It is expressed practically by grams, volts, pounds 
or other unit. 

Resistance is a counter-force or whatever opposes 
the action of a force. 

Work is force exercised in traversing a space 
against a resistance or counter-force. Force multi- 
plied by space denotes work as foot-pounds. 

Energy is the capacity for doing work and is 
measurable by the work units. 

Mass is quantity of matter. 

Weight is the force apparent when gravity acts 
upon mass. When the latter is prevented from 
moving under the stress of gravity its weight can 
be appreciated. 



12 ARITHMETIC OF ELECTRICITY. 

thousand times, or mega, one million times are used, 
as dekalitre, ten liters, kilowatt, one thousand watts, 
megohm, one million ohms. 

Sometimes, where smaller units are wanted, the 
prefixes, deci, one tenth, centi, one hundredth, milli, 
one thousandth, micro, one millionth, are used. A 
microfarad is one millionth of a farad. 

For the concrete conception of the principal units 
the following data are submitted. 

A DanielFs battery maintains an E. M. P. of 1.07 
volt. A current which in each second deposits 
.00033 grams copper (by electro-plating) is of one 
ampere intensity and from what has been said the 
copper deposited by that current in one second cor- 
responds to one coulomb. A column of mercury one 
millimeter square and 106.24 centimeters long has 
a resistance of one ohm at 0° C. The capacity of the 
earth is r oUlo o farad. A Leyden jar with a total 
coated surface of one square meter and glass one 
mm. thick has a capacity of sV microfarad. The 
last is the more generally used unit of capacity. 

These practical units are derived from the O. G. 
S. units by substituting for the centimeter (C.) one 
thousand million (10 9 ) centimeters and for the gram, 
the one hundred thousand millionth (10" 11 ) part of a 
gram. 



CHAPTER II. 

ohm's law. 

This law expresses the relation in an active 
electric circuit (circuit through which a current of 
electricity is forced) of current, electromotive force, 
and resistance. These three factors are always pres- 
ent in such a circuit. Its general statement is as 
follows : 

In an active electric circuit the current is equal to 
the electromotive force divided by the resistance. 

This law can be expressed in various ways as it is 
transposed. It may be given as a group of rules, to 
be referred to under the general title of ohm's law. 

Rule 1. The current is equal to the electromotive 
force divided by the resistance. c = — 

R 

Rule 2. The electromotive force is equal to the cur- 
rent multiplied by the resistance. E = C R 

Rule 3. The resistance is equal to the electromotive 
force divided by the current. r = — 

C 

Rule 4. The current varies directly with the electro* 
motive force and inversely with the resistance* 



14 ARITHMETIC OF ELECTRICITY. 

Rule 5. The resistance varies directly with the elec- 
tromotive force and inversely with the current. 

Rule 6. The electromotive force varies directly with 
the current and with the resistance. 

This law is the fundamental principle in most 
electric calculations. If thoroughly understood it 
will apply in some shape to almost all engineering 
problems. The forms 1, 2, and 3 are applicable to 
integral or single conductor circuits; when two or 
more circuits are to be compared the 4th, 5th and 6th 
are useful. The law will be illustrated by examples. 

Single Conductor Closed Cikcuits. 

These are circuits embracing a continuous con* 
ducting path with a source of electromotive force 
included in it and hence with a current continually 
circulating through them. 

Examples. 

A battery of resistance 3 ohms and E. M. F. 1.07 
volts sends a current through a line of wire of 55 
ohms resistance ; what is the current? 

Solution : The resistance is 3 + 55 = 58 ohms. 
By rule 1 we have for the current 1: ii- giving .01845 
Ampere. 

Note. — A point to be noticed here is that whatever is 
included in a circuit forms a portion of it and its resistance 
must be included therein. Hence the resistance of the 
battery has to be taken into account. The resistance of a 
battery or generator is sometimes called internal resistance 



0HM J 8 LAW. 15 

to distinguish it from the resistance of the outer circuit, 
called external resistance. Resistance in general is 
denoted by R, electromotive force by E, and current by 
C. 

A battery of K 2 ohms; sends a current of .035 
ampere through a wire of R 48 ohms; what is the 
E. M. P. of the battery? 

Solution: The resistance is 48 + 2 = 50 ohms. By 
Eule 2 we have as the E. M. P. 50 X .035 = 1.75 
volts. 

A maximum difference of potential E. M. F. of 
30 volts is maintained in a circuit and a current of 
191 amperes is the result; what is the resistance of 
the circuit? 

Solution: By Rule 3 the resistance is equal to 
iWr=. 157 ohms. 

In the same circuit several generators or gal- 
vanic couples may be included., some opposing the 
others, i. e. connected in opposition. All such can 
be conceived of as arranged in two sets, distrib- 
uted according to the direction of current produced 
by the constituent elements, in other words, so as 
to put together all the generators of like polarity. 
The voltages of each set are to be added together to 
get the total E. M. F. of each set. 

Rule 7. Where batteries or generators are In opposi- 
tion, add together the E. OT. F of all generators of like 
polarity, thus obtaining two opposed E. M. F.s. Sub- 
tract the smaller E. M. F. from the larger E. Itt. F. to 



16 



ARITHMETIC OF ELECTRICITY. 



obtain the effective E. M. F, Then apply Ohm's law on 
this basis of E. ML. F. 

It will be understood that the resistances of all 
batteries or generators in series are added to give 
the internal resistance. 

Examples. 
There are four batteries in a circuit: Battery No. 
1 of 2 volts, }4 ohm;' Battery No. 2 of 1.75 volts, 2 
ohms; Battery No. 3 of 1 volt, 1 ohm; Battery No. 
4 of 1 volt, 4 ohms constants; Batteries 1 and 4 are 
in opposition to 2 and 3. What are the effective 
battery constants? 




-A/WV\AAA/\A^ 



Solution: Voltage = (2 + 1) — (1.75 + 1) = .25 volt. 
Eesistance = j4 + 2+1 + 4 = 7^4 ohms, or .25 
volt, 7^ ohms constants. 

What current will such a combination produce in 
a circuit of 5 ohms resistance? 

Solution: By Ohm's law, Rule 1, the current = 
.25 ■*- (7% + 5) = .02 amperes. 

A battery of 51 volts E. M. F. and 20 ohms resist- 



OHM'S LAW. 17 

ance has opposed to it in the same circuit a battery 
of 26 volts E. M. F. and 25 ohms resistance. A 
current of y& ampere is maintained in the circuit. 
What is the resistance of the wire leads and con- 
nections? 

Solution: The effective E. M. F. is 51 — 26 = 25 
volts. By Rule 3 we have 25 ■*■ i = 200 ohms, as the 
total resistance. But the resistance of the batteries 
(internal resistance) is 20 + 25 = 45 ohms. The re- 
sistance of leads, etc. (external resistance), is there- 
fore 200 — 45 = 155 ohms. 

Portions of Circuits. 

All portions of a circuit receive the same current, 
but the E. M. F., in this case termed preferably dif- 
ference of potential, or drop or fall of potential, 
and the resistance may vary to any extent in differ- 
ent sections or fractions of the circuit. Ohm's Law 
applies to these cases also. 

Examples. 

An electric generator of unknown resistance main- 
tains a difference of potential of 10 volts between its 
terminals connected as described. The terminals 
are connected to and the circuit is closed through a 
series of three coils, one of 100 ohms, one of 50 
ohms, and one of 25 ohms resistance. The connec- 
tions between these parts are of negligibly low re- 



18 ARITHMETIC OF ELECTRICITY. 

sistance. What difference of potential exists be- 
tween the two terminals of each coil respectively? 

Solution: The solution is most clearly reached by 
a statement of the proportion expressed in Rule 6, 
viz. : The electromotive force varies directly with the 
resistance. The resistance of the three coils is 175 
ohms; calling them 1, 2, and 3, and their differences 
of potential E 1 , E 2 , and E 3 , we have the continued 
proportion, 175 : 100 : 50 : 25 :: 10 volts : E 1 : E 2 : E 3 . 
because by the conditions of the problem the total 
E. M. F. ■= 10. Solving the proportion by the regu- 
lar rule, we find that E 1 — 5.7, E 3 — 2.8 and E 3 — 1.4 
volts. 

The same external circuit is connected to a 
battery of 30 ohms resistance. The difference of 
potential of the 100 ohm coil is found to be 30 volts. 
What is the difference of potential between the ter- 
minals of the battery, and what is the E. M. F. of 
the battery on open circuit, known as its voltage or 
E. M. F. (one of the battery constants)? 

Solution: The total external resistance is 100 + 50 
+ 25 = 175 ohms. By Rule 6, we have 100 : 175 :: 30 
volts : x «= 52)4 volts, difference of potential between 
the terminals of the battery. The current is found 
by dividing (Rule 1), the difference of potential of 
the 100 ohm coil by its resistance. This E. M. F. is 
30. The current therefore is t% amperes. The to- 
tal resistance of the circuit is that of the three coils 
or 175 ohms plies that of the battery or 30 ohms, a 



OHM'S LAW. 19 

total of 205 ohms. To maintain a current of rfft 
amperes through 205 ohms (Rule 2), an E. M. F. is 
required equal to A * X 205 volts or 61^ volts. 

Divided, Branched or Shunt Circuits. 

A single conductor, from one terminal of a gener- 
ator may be divided into one or more branches 
which may reunite before reaching the other ter- 
minal. Such branches may vary widely in resist- 
ance. 

Rule 8. In divided circuits, each branch passes a 
portion of a current inversely proportional to its re- 
sistance. 

Examples. 

A portion of a circuit consists of two conductors, 
A and B, in parallel of A =* 50, and B = 75 ohms, 
respectively; what will be the ratio of the currents 
passing through the circuit, which will go through 
each conductor? 

Solution: The ratio will be current through A : 
current through B :: 75 : 50, which may be ex- 
pressed fractionally, *V : ?V. 

Where more than two resistances are in parallel, 
the fractional method is most easily applied. 

Three conductors of A ■=■ 25, B *=» 50, and = 75 
ohms are in parallel. What will be the ratio of cur- 
rents passing through each one? 

Solution: Fractionally A : B : G :: & : & : fa. 



20 ARITHMETIC OE ELECTRICITY. 

Rule 9. To determine the amount of a given cur- 
rent Hi at will pass through parallel circuits of differ- 
ent resistances, proceed as follows : Take the resist- 
ance of each branch for a denominator of a fraction 
having 1 for its numerator. In other words, for each 
branch write down the reciprocal of its resistance. 
Then reduce the fractions to a common denominator, 
and add together the numerators. Taking this sum 
of the numerators for a new common denominator, 
and the original single numerators as numerators, 
the new fractions will express the proportional cur- 
rents as fractions of one. If the total amperage is 
given, it is to be multiplied by the fractions to give 
the amperes passed by each branch. The solution can 
also be done in decimals. 



Examples. 

A lead of wire divides into three branches; No. 
1 has a resistance of 10,000 ohms, No. 2 of 39 ohms, 
and No. 3 of i ohm. They unite at one point. 
What proportion of a unitary current will pass each 
branch? 

Solution: The proportion of currents passed are 
as tt4™ : A : % or 3. Reducing to a common de- 
nominator, these become -s aSoofi : sWtt% : Winnn?-. The 
proportions of the numerators is the one sought for; 
taking the sum of the numerators as a common de- 
nominator, we have in common fractions the follow- 
ing proportions of any current passed by the three 
branches. No. 1, tt&W; No. 2, jiiUh; No. 3, 

iwm. 

Four parallel members of a circuit have resistances 
respectively of 25, 85, 90, and 175 ohms; express 



OHM'S LAW. 21 

decimally the ratio of a unitary current that will 
pass through them. 

Solution: The ratio is as & : -fa : A : xfr, or reduc- 
ing to decimals (best by logarithms), .04 : .011765 : 
.011111 : .0057. Adding these together, we have 
.068576, which must be multiplied by 14.582 to pro- 
duce unity. Multiplying each decimal by 14.58 
(best by logarithms), we get the unitary ratio as 
.5832 : .17153 : .1620 : .08310, whose sum is 1.0000. 

Unless logarithms are used, it is far better to work 
by vulgar fractions. 

A current of .71 amperes passes through two 
branches of a circuit. One is a lamp with its con- 
nections of 115 ohms resistance; another is a resist- 
ance coil of 275 ohms resistance. What current 
passes through each branch? 

Solution: The proportions of the current are as 
rh : sh or reduced to a common denominator and to 
their lowest terms drk : *H*. Proceeding as before, 
and taking the sum of the numerators (55 + 23 — 
78), as a common denominator, we find that the lamp 
passes H, and the resistance coil H of the whole 
current. Multiplying the whole current, .71 by W, 
we get H%$ amperes, or i ampere for the lamp, leaving 
.21 or a little over i ampere for the resistance coil. 

Another problem in connection with parallel 
branches of a circuit is the combined resistance of 
parallel circuits. This is not a case of summa- 



22 ARITHMETIC OF ELECTRICITY. 

tion, for it is evident that the more parallel paths 
there are provided for the current, the less will be 
the resistance. 

Rule 10. In shunt circuits, the resistance of the com- 
bined shunts is expressed by the reciprocal of the sum 
of the reciprocals of the resistances. 

Example. 

Two leads of a 50 volt circuit (leads differing in 
potential by 50 volts), are connected by a 20 ohm 
motor. A 50 ohm lamp and 100Q ohm resistance 
coil are connected in parallel or shunt circuit there- 
with, what is the combined resistance? and the total 
current? 

Solution: The reciprocal of resistance is conduc- 
tance, sometimes expressed as mhos. (Rule 19.) 
The conductance of the three shunts is equal to 
A + A + tAtf mhos = rlHb + TrHfo + tAt = rbfo 
mhos. The reciprocal of conductance is resistance. 
The combined resistance is therefore -HP" ohms = 
14.09 ohms. The current is J£L or 3.5 amperes. 

Rule 11. The combined resistance of two parallel 
circuits is found by multiplying the resistances to- 
gether, and dividing the product by the sum of the re- 
sistances. Where there are several circuits, any two 
can be treated thus, and the result combined in the 
same way with another circuit, and so on to get the 
final resistance. _, _ rXrl 



R = 



r+r» 



Example. 

Four conductors in parallel have resistances of 
100 — 50 — 27 — 19 ohms. What is their combined 
resistance? 



OHM'S LAW. 23 

Solution: Combining the first and second, we 
have jjjjprijj = 33i ohms. Combining this with the 
resistance of the third wire, we have §jqpff = 14.9 
ohms. Combining this with the resistance of the 
fourth wire, we have u.q + iq "" 8 ' 3 ohms - The 
result is, of course, identical by whatever rule 
obtained. 

Rule 12. When all the parallel circuits are of uni- 
form resistance, as in multiple arc incandescent light- 
ing, the resistance of the combined circuits is found by 
dividing the resistance of one circuit by the number of 

circuits. __ * 

R ~~ n 

Examples. 

There are fifty lamps of 100 ohms resistance each 
in multiple arc connection. What is their com- 
bined resistance? 

Solution: W = 2 ohms. 

A motor can take 3 amperes of currents at 
30 volts safely without burning out or heating 
injuriously. A 110 volt incandescent circuit is at 
hand. The motor is to be connected across the 
leads so as to receive the above amperage. A shunt 
or branch of some resistance is carried around it, 
and a resistance coil intervenes between the united 
branches and one of the main leads. The resistance 
of the coil is 20 ohms. What should the resistance 
of the shunt be? 



24 



ARITHMETIC OF ELECTRICITY. 



31 






•--^vwwvvw 



4° °f 
MM 



Solution: The resistance of the motor (Ohm's 
Law, Rule 3), is found by dividing the E. M. F. by 
the resistance — 30 -*- 3 = 10 ohms. By Rule 5 the 
resistance of the coil in series (20 ohms) must be to 
the combined (not added) resistance of the motor 
and shunt coil, as 110 — 30 (total voltage minus 
voltage for motor) : 30 (voltage for motor) or 20 : 
x :: 80 : 30 . \ x = 7.5 combined resistance of parallel 
or shunt coil and motor. The reciprocal of 7.5 
(conductance, Rule 19), may be expressed as flftths of 
the combined (in this case added) conductances of 
shunt coil and motor. The conductance of the 
motor is equal to the reciprocal of 10 which may be 
expressed as A or as •$$$. The conductance of the 
shunt coil must therefore be H% ~~" Wfr = t¥tt = A 
mho. The reciprocal of this gives the resistance of 
the shunt coil which is 30 ohms. The total current 
going through the system by Ohm's law is 7i + 20 = 4 
amperes. The resistance of the shunt coil — 30 ohms 
—is to that of the motor in parallel with it— 10 
ohms — as the current received by the motor is to 



OHM'S LAW. 26 

that received by the coil, a ratio of 30 : 10 or 3:1 
giving 3 amperes for the motor and 1 ampere for 
the coil. This is a proof of the correctness of 
operations. 

Two conductors through which a current is 
passing are in parallel circuit with each other. 
One has a resistance of 600 ohms. The other has a 
resistance of 3 ohms. A wire is carried across from 
an intermediate point of one to a corresponding 
point of the other. It is attached at such a point of 
the first wire that there are 400 ohms resistance be- 
fore it and 200 after it. Where must it be connected 
to the other in order that no current may pass? 

Solution: The E. M. F. up to the point of con- 
nection of the bridge or cross wire is to the total 
E. M. F. in the 600 ohm wire as 400: 600 or as 2: 3. 
The other wire which by the conditions has the 
same drop of potential in its full length must be 
divided therefore in this ratio. The bridge wire 
must therefore connect at 2 ohms from its begin- 
ning, leaving 1 ohm to follow. The principle here 
illustrated can be proved generally and is the Wheat- 
stone Bridge principle. 



CHAPTER III. 
resistance and conductance. 

Resistance of Different Conductors of the 
same Material. 

Conductors are generally circular in section. 
Hence they vary in section with the square of their 
diameters. The rule for the resistance of conduc- 
tors is as follows: 

Rule 13. The resistance of conductors of identical 
material varies inversely as their section, or if of circu- 
lar section inversely as the squares of their diameters, 
and directly as their lengths. 

Example. 

1. A wire a, is 30 mils in diameter and 320 feet 
long; another b, is 28 mils in diameter and 315 feet 
long. What are their relative resistances? 

Solution: Calling the resistances R a : E b we would 
have the inverse proportion if they were of equal 
lengths E b : R a :: SO 2 : 28* or as 900 : 784. Were 
they of equal diameter the direct proportion would 
hold for their lengths: R b : R a :: 315 : 320. Com- 
bining the two by multiplication we have the com- 
pound proportion R b : R a :: 900 X 315 : 784 X 320 or 
as 283,500 : 250,880, or as 28 : 25 nearly. The 



RESISTANCE AND CONDUCTANCE. 27 

combined proportions could have been originally 
expressed as a compound proportion thus: R b : R a :: 
30 2 X 315 : 28 2 X 320. 

For wires of equal resistance the following is 
given. 

Rule 14. The length of one wire multiplied by the 
square of the diameter of the other wire must equal the 
square of its own diameter multiplied by the length of 
the other if their resistances are equal. Or multiply the 
length of the first wire by the square of the diameter of 
the second. This divided by the length of the second 
will give the square of the diameter of the first wire; 
or divided by the square of the diameter of the first will 
give the length of the second. Id 73 — I'd? 

Examples. 

1. There are three wires, a is 2 mils, b is 3 mils, 
and c is 4 mils in diameter; what length must b 
and c have to be equal in resistance to ten feet of a? 

Solution: Take a and c first and apply the rule, 
10 X 4 a -+-2 2 = 40 feet; then take a and b 10 X 
3 2 ■*■ 2 2 = 22% feet. To prove it compare a and c 
directly by the same rule 22% X 4 2 + 3 2 = 40. 
As this gives the same result as the first operation, 
we may regard it as proved. 

A conductor is 75 mils in diameter and 79 feet 
long; how thick must a wire 1264 feet long be to 
equal it in resistance? 

Solution: 75 2 X 1264 + 79 = 7,110,000 + 79 = 
90,000. The square root of this amount is 300 which 
is the required diameter. 



28 ARITHMETIC OF ELECTRICITY. 

For problems involving the comparison of wires of 
unequal resistance the rule may be thus stated: 

Rule 15. Multiply the square of the diameter of each 
wire by the length of the other. Of the two products 
divide the one by the other to get the ratio of resist- 
ance of the dividend to that of the divisor taken at 
unity. The term including the length of a given wire 
is the one expressing the relative resistance of such 
wire. 

Examples. 

A wire is 40 mils in diameter, 3 miles long and 
40 ohms resistance. A second wire is 50 mils in 
diameter and 9 miles long. What is its resistance? 

Solution: 9 X 40 2 = 14,400 relative resistance of 
the first wire. 3 X 50 2 = 7,500 relative resistance 
of second wire. 14,400 ■#■ 7,500 = 1.92 — ratio of 
resistance of second wire to that of first taken at 
unity. But the latter resistance really is 40 ohms. 
Therefore the resistance of the second wire is 40 X 
1.92 = 76.80 ohms. 

The result may also be worked out thus: 

40 2 X 9 = 14,400 = relative resistance of the 3 mile 
wire. 

50 2 X 3 = 75Q0 « re i a tive resistance of the 9 mile 
wire. 

14,400 -*- 7500 = 1.92 = ratio of 9 mile (dividend) 
to 3 mile (divisor) wire. 

,\ 40 ohms X 1.92 = 76.8 ohms. 

A length of a thousand feet of wire 95 mils in 
diameter has 1.15 ohms resistance; what is the di- 



RESISTANCE AND CONDUCTANCE. 29 

ameter of a wire of the same material of which the 
resistance of 1000 feet is 10.09 ohms? (R. E. Day, 
M. A.). 

Solution: 10.09 ■+■ 1.15 = 8.77 ratio of resistances. 
If we divide 1000 by 8. 77 we obtain a length of the 
first wire which reduces the question to one of iden- 
tical resistances. 1000 -*■ 877 = 114 feet. Then 
applying Rule 14, 114 X 952 ■*- 1000 = 1037.88. 
This is the square of the diameter of the other wire. 
Its square root gives the answer: 32.2 mils. 

Specific Resistance. 

Specific resistance is the resistance of a cube of 
one centimeter diameter of the substance in ques- 
tion between opposite sides. It is expressed in 
ohms for solutions and in microhms for metals. 
From it may be determined the resistance of all 
volumes, generally prisms or cylinders, of substance. 
Very full tables of Specific Resistance are given in 
their place. 

Rule 16. The resistance of any prism or cylinder of a 
substance Is equal to Its specific resistance multiplied. 
by its length in centimeters and divided by its cross- 
sectional area in square centimeters. If the dimensions 
are given in inches or other units of measurements 
they must be reduced to centimeters by the table. 

R = Sp. R x I 
a 

Examples. 
An electro-plater has a bath of sulphate of copper, 
gp. resistance 40 ohms. His electrodes are each J 



30 ABITHMETIG OF ELECTRICITY. 

foot square and 1 foot apart. What is the resist- 
ance of such a bath ? 

Solution : By the table 1 square foot = 929 sq. cent, 
and 1 foot = 30.4797 cent. .-. Eesistance = 40 X 
30.4797 + 929 = 1.31 ohms. 

Where the electrodes in a solution are of uneven 
size take their average size per area. The facing 
areas are usually the only ones calculated, as owing 
to polarization the rear faces are of slight efficiency, 
and where the electrodes are nearly as wide ae the 
bath or cell the active prism is practically of cross- 
sectional area equal to the area of one side of a plate. 

In a Bunsen battery the specific resistances of the 
solutions in inner and outer cells were made alike, 
each equalling 9 ohms. The central element was a 
y 2 inch cylinder of electric light carbon. The outer 
element was a plate of zinc 6 inches long bent into a 
circle. When there were 2 inches of solution in the 
cell what was the resistance? 

Solution : Area of carbon = ?X2= 3.14 square 
inches. Area of zino = 2 X 6 = 12 square inches. 
This gives an average facing area of (12 + 3.14) ■*■ 
2 = 7.57 square inches = 48.38 sq. cent. The 
distance apart = ^ inches (nearly) = 1.9049 cent. 
.♦. Eesistance = 9 X 1.9049 *- 48.38 = .354 ohms. 

For wires, the specific resistance of metals being 
given in microhms, the calculation may be made in 
microhms, or in ohms directly. As wire is cylindri- 



RESISTANCE AND CONDUCTANCE. 31 

cal a special calculation may be made in its case to 
reduce area of cross section to diameter. This may 
readily be taken from the table of wire factors, thus 
avoiding all calculation. 

Rule 17. The resistance in microhms of a wire of given 
diameter in centimeters is equal to the product of the 
specific resistance by 1.2737 by the length in centi- 
meters divided by the square of the diameter in cen- 
timeters. 

_ Sp. Res, x '.2737 x I 

< — d» " 

Examples. 

The Sp. Kes. of copper being taken at 1.652 mi- 
crohms what is the resistance of a meter and a half 
of copper wire 1 millimeter thick? 

Solution: The diameter of the wire (1 millimeter) 
is .1 centimeter. The square of .1 is .01. The 
length of the wire (1*4 meter) is 150 centimeters. 
Its resistance therefore is 1.652 X 1.2737 X 150 -*■ 
.01 = 31,561 microhms or .031,561 ohms. 

Universal Eule for Kesistances. 

Into the problem of resistances of one or two wires 
eight factors can enter, these are the lengths, sec- 
tional areas, specific resistances and absolute resist- 
ances of two wires. Their relation may be ex- 
pressed by an algebraic equation, which by transposi- 
tion may be made to fit any case. The rule is 
arithmetically expressed by adopting the method of 
cancellation, drawing a vertical line and placing on 



32 ARITHMETIC OF ELECTRICITY. 

the left side, factors to be multiplied together for 
a divisor, and on the right side factors to be multi- 
plied together for a dividend. In the expression of 
the rule as below the quotient is 1, in other words 
the product of all the factors on the left hand of 
the line is equal to that of all the factors on the 
right hand. Calling one wire a and the other b we 
have the following expression: 



Kesistance of b 
Specific Eesistance of a 
Length of a 
Cross-sectional area of b 



Eesistance of a 
Specific Eesistance of b 
Length of b 
Cross-sectional area of d 



Rule 1 8 • Substitute in the above expression the values 
of any factors given. Substitute for factors not given 
or required the figure 1 or unity. Such a value deter- 
mined by division must be given to the required factor 
and substituted in its place as will make the product 
of the left-hand factors equal to that of the right-hand 
factors. Only one factor can be determined, and all 
factors not given are assumed to be respectively equal 
for both conductors. 

Examples. 

If the resistance of 500 feet of a certain wire is 
,09 ohms what is the resistance of 1050 feet of the 
same wire? 

Solution: The cross sectional areas and specific 
resistance not being given are taken as equal. (This 
of course follows from the identical wire being re- 
ferred to.) The vertical line is drawn and the 
values substituted : 



RESISTANCE AND CONDUCTANCE. 33 

09 



Resistances : Eesistance of 
required wire 



1050 



Lengths : 500 

(Other factors omitted as unnecessary.) 
1050 X .09 -*- 500 = .189 ohms. 

What is the diameter of a wire 2 miles long of 
23 ohms resistance, if a mile of wire of similar ma- 
terial of seventy mils diameter has a resistance of 
10.82 ohms? 

Solution. We use for simplicity the square of the 
diameter in place of the cross sectional area of the 
known wire, thus: 



Resistances : 23 

Lengths : 1 

Areas : Unknown 



10.82 

2 
70 2 



As the specific resistances are identical they are 
not given. 

2 X 70 2 X 10.82 + 23 X 1 = 4610 square of diameter 
required : 4610^ = 68 mils. 

What must be the length of an iron wire of cross- 
sectional area 4 square millimeters to have the same 
resistance as a wire of pure copper 1000 yards long, 
of cross-sectional area 1 square millimeter, taking 
the conductance of iron as \ that of copper? (Day). 

Solution: 



34 



ARITHMETIC OF ELECTRICITY. 



Specific Resistances : 1 

Lengths : 1000 

Cross-sectional areas : 4 



7 (i.e. the reciprocal of 

conductance) 
Unknown 



As the resistances are identical they are not 
given. 

Solving we have 1000 X4+7 = 571$ yards. 

There are two conductors, one of 35 ohms resist- 
ance, 1728 feet long and 12 square millimetres cross- 
sectional area and specific resistance 7: the other of 
14 ohms resistance, 432 feet long and 8 square milli- 
metres cross-sectional area. What is its specific 
resistance? 



Resistances : 35 

Specific Resisianoes 
UnJcnotvn 
Lengths : 432 

Cross- Sectional areas : 12 



14 

7 

1728 

8 



By cancellation this reduces tol4X8-*-5X3 = 
7.4 Specific Eesistance. 

In these cases it is well to call one wire a and the 
other by and to arrange the given factors in two 
columns headed by these designations. Then the 
formula can be applied with less chance of error. 
Thus for the last two problems the columns should 
be thus arranged. 



BESISTANCE AND CONDUCTANCE. 



35 



Area, 4 sq. mils. 
L. Unknown 
Sp. Res. 7 



1,000 yards 

1 



Resist. 35 ohms 
L. 1,728 feet 
Area, 12 sq. mils. 
Sp. Res 7 



14 ohms 
432 feet 
8 sq. mils, 
required 



From such statements of known data the formula 
can be conveniently filled up. 

Eesistakce of Wires Eeferred to Weight. 

The weight of equal lengths of wire is in propor- 
tion to their sections. The problems involving 
weight therefore can be reduced to the Eules al- 
ready given. 

Problem. A wire, A, is 334 feet long and weighs 25 
oz.; another, B, is 20 feet long and weighs 1 oz. 
what are the relative resistances? 

Solution: 20 feet of the wire "A" weigh ^ x 
25 = 1.50 oz. The weights of equal lengths of A 
and B respectively areas 1.50 : 1.00 which is also 
the inverse ratio of the resistances of equal lengths. 
By compound proportion Eule we have E. of " A " 
: E. of "B":: 1 X 334 :: 1.50 X 20; reducing to 
16.7 : 1.5 or 11.1 : 1.0 or the wire " A " has about 
eleven times the resistance of the wire " B." 

Solution: By general Eule for resistance (Eule 
18). Taking 1.50 : 1.00 as the ratio of cross-sec- 
tional areas and taking the resistance of the long 
wire A as 1 we have : 



36 ARITHMETIC OF ELECTRICITY. 



Resistances : 1 

Lengths : 20 

Cross-sectional area : 1.50 



Unknown 

334 

1 



Eesistance of B = 1.50 X 20 ■*■ 334 = .0899 or 
about A as before. 

Conductance. 

Conductance is the reciprocal of resistance and is 
sometimes expressed in units called mhos, which 
is derived from the word ohm written backwards. 

Rule 19. To reduce resistance in ohms to conductance 
in nilios express its reciprocal and the reverse* 

K = R 

Examples. 

A wire has a resistance of Us ohms, what is its 
conductance? 

Solution: 126 •*- 18 = 7 mhos. 

Reduce a conductance of lif to ohms. 
Solution: lit = ft mhos which gives if ohms. 

It is evident that the data for problems or that 
constants could be given in mhos instead of ohms. 
In some ways it is to be regretted that the positive 
quality of conductance was not adopted at the out- 
set instead of the negative quality of resistance. 
One or two illustrations may be given in the form of 
examples involving conductance. 

Express Ohm's law in its three first forms in 
conductance. 



RESISTANCE AND CONDUCTANCE. 37 

Solution: This is done by replacing the factor 
u resistance " by its reciprocal. Thus, Rule 1 
reads for conductance: " The current is equal to 
the electromotive force multiplied by the conduc- 
tance" (C = EK)— Rule 2 as "The electromotive 
force is equal to the current divided by the conduc- 
tance" (E = — ) — Rule 3 as "The conductance is 

K 
equal to the current divided by the electromotive 


force." (K = — ) 

E 

A circuit has a resistance of .5 ohm and an E. M. 
F. of 50 volts; determine the current, using con- 
ductance method. 

Solution: The conductance = ~ = 2 mhos. The 
current = 50 X 2 = 100 amperes. 

In a circuit a current of 20 amperes is main- 
tained through 2f ohms. Determine the E. M. P. 
using conductance. 

Solution: The conductance = 4 = 4$ mhos. 
E. M. F = 20 + i$ = 52 volts. 

Assume a current of 30 amperes and an E. M. 
P. of 50 volts, what is the conductance and resis- 
tance? 

Solution: Conductance = 30 -*- 50 = .6 mho. 
Resistance =1 -*- .6 = 1.667. 



CHAPTER IV. 

potential difference. 

Drop of Potential in Leads and Size of 
Same for Multiple Arc Connections. 

Subsidiary leads are leads taken from large sized 
mains of constant E. M. P. or from terminals 
of constant E. M. F. to supply one or more lamps, 
motors, or other appliances. A constant voltage is 
maintained in the mains or terminals. There is a 
drop of potential in the leads so that the appliances 
always have to work at a diminished E. M. P. The 
E. M. P. of the leads is known, the requisite 
E. M. P. and resistance of the appliance is known, 
a rule is required to calculate the size of the wire to 
secure the proper results. It is based on the princi- 
ple that the drop or fall in potential in portions of 
integral circuits varies with the resistance. (See 
Ohm's law). A rule is required for a single appli- 
ance or for several connected in parallel. 

Rule 20. The resistance of the leads supplying any ap- 
pliance or appliances for a desired drop in potential 
within the leads is equal to the reciprocal of the cur- 
rent of the appliances multiplied by the desired drop 
in volts. 



POTENTIAL DIFFERENCE. 39 

Example. 

A lamp, 100 volts X 200 ohms, is placed 100 feet 
from the mains, in which mains a constant E. M. F. 
of 110 volts is maintained. What must be the resist- 
ance of the line per foot of its length; and what 
size copper wire must be used? 

Solution: The lamp current is obtained (Ohm's 
law) by dividing its voltage by its resistance, (M# = 
| ampere). The reciprocal of the current is f ; mul- 
tiplied by the drop (f X 10 = 20) it gives the resist- 
ance of the line as 20 ohms. As the lamp is 100 
feet from the mains there are 200 feet of the wire. 
Its resistance per foot is therefore -gfo = A ohm or 
it is No. 30 A. W. GL (about). 

For several appliances in parallel on two leads a 
similar rule may be applied. There is inevitably a 
variation in E. M. F. supplied to the different appli- 
ances unless resistances are intercalated between the 
appliances and the leads. 

Rule 21. The E. M. F. of the main leads or terminals 
the factors of the lamps or other appliances, their num. 
ber and the distance of their point of connection are 
given. The combined resistance is found by Rules 8 to 
12. Then by Rule 20 the resistance of the leads is cal- 
culated. 

Example. 

A pair of house leads includes 260 feet of wire, 
or 130 in each lead. Six 50 volt 100 ohm lamps are 
connected thereto at the ends. The drop is to be 5 



40 ARITHMETIC OF ELECTRICITY. 

volts, giving 55 volts in the main leads. Kequired the 
total resistance of and size of wire for the house leads. 
Solution: The resistance of six 100 ohm lamps in 
parallel is 100 ■*■ 6 = 16.66 ohms. The current re- 
quired is by Ohm's law 50 -s- 16.66 or 3 amperes. 
Its reciprocal multiplied by the drop, (1X5 = 1 = 
\ 2 /z ohms) gives the required resistance = \ 2 /z ohms. 
This, divided by 260 feet gives the resistance per 
foot as ,0064 ohm, corresponding by the table to 
No. 18 A. W. G. 

A rule for the above cases is sometimes expressed 
otherwise, being based on the proportion: Eesist- 
ance of appliances is to resistance of leads as 100 
minus the drop expressed as a percentage is to the 
drop expressed as a percentage. This gives the fol- 
lowing: 

Rule 22. The resistance of the leads Is equal to the 
combined resistance of the appliances multiplied by 
the percentage of drop and divided by 100 minus the 
percentage of drop. 

Problem. Take the data of last problem and 
solve. 

Solution: The percentage of drop is A = 9g. The 
resistance of the leads = 16 - 66 * 9 = 1^3 = IV3 ohms 

100 — 9 91 ' 

about. 

Note. — To obtain accurate results the figures of percen- 
tage, etc., must be carried out to two or more decimal 
places. Rules 20 and 21 are to be preferred to any percen- 
tage rule. Also see Rule 23. 



POTENTIAL DIFFERENCE. 41 

Where groups of lamps are to be connected along 
a pair of leads but at considerable intervals, the 
succeeding sections of leads have to be of diminish- 
ing size. The same problem arises in calculating 
the sizes of street leads. The identical rule is ap- 
plied, care being taken to express correctly the ex- 
act current going through each section of the lead. 
The calculation is begun at the outer end of the 
leads. A diagram is very convenient; it may be 
conventional as shown below. 

Examples. 

At three points on a pair of mains three groups of 
fifty 220 ohm lamps in parallel are connected; a 
total drop of 5 volts is to be divided among the 
three groups, thus: 1.6 volts — 1.6 volts — 1.8 volts. 
The initial E. M. F. is 115 volts; what must be the 
resistances of the three sections of wire? 

Solution: The following diagram gives the data 
as detailed above: 

1. 2. 3. 

S&JZajty* S&JUtmpa. Si?-L<un./>s 

-.X*-WbJk*/>.- -Z6KtlsJDrep.~ ~/.£KlZs7hvp.- 

Starting at group 3 we have 50 lamps in parallel 
each of 220 ohms resistance, giving a combined re- 
sistance (Rule 12) of 4.4 ohms and a total current 
(Ohm's law) of 110 ■*■ 4.4 == 25 amperes. The re- 
sistance of section 2 — 3 is by the present rule ■£% X 



42 



ARITHMETIC OF ELECTRICITY. 



1.8 = .072 ohms. Taking group 2 the current 
through this group of lamps is 111.8 -*- 4.4 = 25.41 
amperes. The section 1 — 2 has to pass also the cur- 
rent 25 amperes for group 3 giving a total current 
of 25 + 25.41 = 50.41 amperes. The resistance of 
section 1 — 2 is therefore ^j 1 X 1.6 = .0317 ohm. 
Taking group 1 the current for its lamps is 113.4 *■ 
4.4 = 25.7 amperes. The total current through 
section 0—1 is therefore 25 + 25.4 + 25.7= 76.1 
amperes. The resistance of the section is ^ X 1.6 
= .021 ohms. Arranging all these data upon a 
diagram we have the full presentation of the calcu- 
lation in brief as below: 



iPjLfJ>Amp A _ 



/0<*/r£Af9*s „_ M \ 



4rrvey»f. 



IT * 



(rrviyod. 



CHAPTER V. 

CIRCULAR MILS. 

A mil is ttAto of an inch. The area of a circle, 
one mil in diameter, is termed a circular mil. The 
area of the cross-section of wires is often expressed in 
circular mils. Thus a wire, 3 mils in diameter, has 
an area of 9 circular mils, as shown in the cut. A 




circular mil is .7854 square mil. Rules for the sizes 
of wires for given resistances are often based on cir- 
cular mils, and include a constant for the conduc- 
tivity of copper. By the table of specific resistances, 
the values found can be reduced to wires of iron or 
other metals. 



44 ARITHMETIC OF ELECTRICITY. 

A commercial copper wire, one foot long, and one 
circular mil in section, has a resistance of 10.79 
ohms at 75° F. This is, of course, largely an as- 
sumption, but is taken as representing a good aver- 
age. No two samples of wire are exactly alike, and 
many vary largely. From Eule 13, and from the 
above constant, we derive the following rules: 

Rule 23. The resistance of a commercial copper wire 
is equal to its length divided by the cross-section in cir- 
cular mils, and multiplied by 10.79. 

Example. 

A wire is 1050 feet long, and has a cross-section 
of 8234 circular mils. What is its resistance? 
Solution: 1050 X 10.79 -*- 8234 = 1.37 ohms. 

Rule 24. The cross-section of a wire in circular mils 
is equal to its length divided by its resistance, and mul- 
tiplied by 10.79. 

Example. 

A wire is 1050 feet long, and has a resistance of 
.68795 ohms. What is its cross-section in circular 
mils? 

Solution: 1050 X 10.79 -*- .68795 = 16,468 circu- 
lar mils. 

Rule 25. The cross-section of the wires of a pair of 
leads in circular mils for a given drop expressed in per- 
centage is equal to the product of the length of leads by 
the number of lamps (in parallel), by 21.58, by the dif- 
ference between 1 00 and the drop, the whole divided 
by the resistance of a single lamp multiplied by th© 
drop. In X 21 .58 X (100- e) 

JJL —— — _— — — _ ^^ 



CIRCULAR MILS. 45 

Example. 

Fifty lamps are to be placed at the end of a double 
lead 150 feet long (= 300 feet of wire). The resist- 
ance of a lamp is 220 ohms. What size must the 
wire be for b% drop? 

Solution: 150 X 50 X 21.58 X (100 — 5) + (220 X 
5) = 13,977.9 circular mils. 

In these calculations and in the calculations given 
on page 48 it is important to bear in mind that the 
percentage is based upon the difference of potential 
at the beginning of the leads or portion thereof 
under consideration; in other words upon the high- 
est difference of potential within the system or the 
portion of the system treated in the calculation 



CHAPTER VL 

special systems. 

Three Wire System. 

As there are three wires in the three wire system, 
where there are two in the ordinary system, and as 
each of the three wires is one quarter the size of 
each of the two ordinary system wires, the copper 
used in the three wire system is three-eighths of 
that used in the ordinary system. 

In the three wire system the lamps are arranged 
in sets of two in series. Hence but one-half the 
current is required. The outer wires, it will be no- 
ticed, have double the potential of the lamps. 
Hence to carry one-half the current with double the 
E. M. F., a wire of one quarter the size used in the 
ordinary system suffices. 

Rule 26. The calculations for plain multiple arc 
work: apply to the three wire system, as regards size of 
each of the three leads, if divided by 4. 

While the central or neutral wire will have noth- 
ing to do when an even number of lamps are burning 
on each side, and may never be worked to its full 
capacity, there is always a chance of its having to 
carry a full current to supply half the lamps (all on 



SPECIAL SYSTEMS. 47 

one side). Hence it is made equal in size to the 
others. 

Alternating Current System. 

The rules already given apply in practice to this 
system also, although theoretically Ohm's law 
and those deduced from it are not correct for this 
current. A calculation has to be made to allow for 
the conversion from primary to secondary current 
in the converter as below. 

Note. — The ratio of primary E. M. F. to secondary is ex- 
pressed by dividing the primary by the secondary, and is 
termed ratio of conversion. Thus in a 1000 volt system with 
50 volt lamps in parallel the ratio of conversion is 1000 -s- 50 
= 20. 

Rule 27. The current in the primary is eqnal to the 
current in the secondary divided toy the ratio of conver- 
sion. 

Examples. 

On an alternating current circuit whose ratio of 
conversion = 20, there are 1000 lamps, each 50 volt; 
50 ohms. When all are lighted what is the primary 
current? 

Solution : By Ohm's law the secondary current is 
1000 x 1 (each lamp using ■£# = 1 ampere, Ohnrs 
law) = 1000 amperes. 1000 -*- 20 = 50 amperes is 
the primary current. 

The current being thus determined the ordinary 
rules all apply exactly as given for direct current 
work. 



48 ARITHMETIC OF ELECTRICITY. 

Given 650 lamps, 50 volt 50 ohms each, 3600 
feet from the station. The primary circuit pressure 
is 1031 volts. A drop of 3% is to be allowed for in 
the primary leads. What is the resistance of the 
primary wire? 

Solution: Current of a single lamp = 50 + 50 = 1 
ampere; current of 650 lamps = 650 amperes, cur- 
rent of primary 650 -*- 20 == 32| amperes, drop of 
primary = 1031 X 3% = 30.9 volts, resistance of pri- 
mary (Kule 20) 1 X 30.9 = .9516 ohm. 

Ilule 28. For obtaining the size of the primary wire 
in circular mils, calculate by Rule 25, and divide the 
result by the square of the ratio of conversion. 

Example. 

Take data of last problem and solve. 
Solution: [3600 X 650 X 21.58 X (100—3) -*- (50 X 
3)] + 20 2 = 81,637 circular mils. 

The two last examples may be made to prove each 
other, thus: 

The total length of leads is 3600 X 2 = 7200 feet. 
If of 1 mil thickness its resistance would be 7200 X 
10.79 = 77,688 ohms. As resistance varies inversely 
as the cross sectional area we have the proportion 

.9516 : 77,688 :: 1 : x which gives x = 81,639 cir- 
cular mils corresponding within limits to the result 
obtained by Eule 28. 

In all cases of this sort where percentage is ex- 
pressed the statement in the last paragraph on page 



SPECIAL SYSTEMS. 49 

45 should be kept in mind. The ratio of conversion 
must be based on the E. M. F. at the coil (in this 
case on 1031 — 31 = 1000 volts) not on the E. M. F. 
at the beginning of the leads or portion thereof con- 
sidered in the calculation. The percentage of drop 
must be subtracted before the ratio of conversion is 
calculated. 

For winding the converters, the following is the 
rule : 

Rule 29. The convolution* of the primary are equal 
in number to the product of the convolutions of the sec- 
ondary multiplied by the ratio of conversion, and vice 
versa* 

Examples. 

The ratio of conversion of a coil is 20; there are 
1000 convolutions of the secondary. How many 
should there be of the primary? 

Solution: 1000 X 20 = 20,000 convolutions. 

There are in a coil 5000 convolutions of the pri- 
mary; its ratio of conversion is 50. How many con- 
volutions should the secondary have? 

Solution: 5000 -*■ 50 = 100 convolutions. 



CHAPTER VIL 

I AND I: 

It fa Joule that 

Aal que, .it is 

equa. :X multiplied bj 

•aw, to the 
square of t: ] by the 

which again i EL M« F. malt 

remom 
ftppli and thv 

KuJ< | ■■■. f h <■ 'T^r-jy or b<rat d#r%»-Jop#:d in <lr«:tift«lfl 

In ;>r ■■■ pn r ti • n t , I h f m^u ar<; of thi ♦: f I rr<: ■ t rrj ul lip J) « J bf 

An ' r ns: it is 

- 

for all | 
in proporti 

The 



61 

heat in tlie i the 

waste is A of the total fa 

internal resistance | to 

. I 
quantiti 
the I 

is wire with the quantities prod 
hen they arc conned 
of the same win . - i /.) 

.tive ci: : in the 

two case the 

K. II F. of the 

unts 
for the resistance of 3 meters of wire, the 
■ ■ : . . v. re thei 

I the 
-d ratio. But ac 
taken into account. ermine 

heat on. B - 

:ning idem! iL Bj Ohm's 

io of J : D pro- 

portion to JJ : T r 5 T = 1"" : 1. 
the proportion of th 

alone, with the short wire h the long 

1. For the heating effects on the outside ci: 

:nce and cuj full for- 

mula of the rule mt; ratio of the 

that in I 



52 ARITHMETIC OF ELECTRICITY. 

wire connection is as (|) 2 X 1 : (£)> X 37 = 100 X 1 : 
37 X 1. The ratio therefore is as 100 is to 37 for 
the total heat produced in the circuit which includes 
battery and connections. 

Owing to irregular working of a dynamo, an in- 
candescent lamp receives sometimes the full amount 
or i ampere of current; at other times as little as H 
ampere. What proportion of heat is developed in it 
in both cases, assuming its resistance to remain 
sensibly the same? 

Solution: By the rule the ratio is (f) : (ii) or i: 
■iwfc = 2025 : 400. The diminution of current there- 
fore cuts down the heat to i the proper amount. 

Rule 31. The energy or heat developed in a circuit is 
in proportion to the square of the electromotive force 
divided by the resistance. E a 

H = - 
II 

Examples. 

There are two Grove batteries, each developing 
1.98 volts E. M. F. One has an internal resistance 
of A ohm; the other of J ohm. They are placed in 
succession on a circuit of 2 ohms resistance. What 
is the ratio of heats developed by the batteries in ' 
each case? 

Solution: As the E. M. F. is constant it may be 
taken as unity. Then for the two cases we have 
8^ : gj ^ 2i : 2A as the ratio of heat produced. 

A battery of one ohm resistance and two volts E. 



WORK AND ENERGY. 53 

M. F. is put in circuit with 4 ohms resistance. 
Another battery of 4 ohms and 1 volt is connected 
through 1 ohm resistance. What ratio of heat is 
developed in each case? 

Solution: ^ : 1*1 or 4 : 1. 

Rule 32. The energy or heat developed in a circuit is 
in proportion to the E. JI. F. multiplied by the current. 

H = EO 

Examples. 

Take data of last problem and solve. 

Solution: For first battery, by Ohm's law, current = 
f ampere; for the second, current = J ampere. The 
heat developed, is by the present rule, in the propor- 
tion as | X 2 : i X 1 or 4 : 1. 

Compare the heat developed in a 100 volt 200 ohm 
lamp and in a 35 volt 35 ohm lamp and in a 50 volt 
50 ohm lamp. 

Solution : The currents (Ohm's law) are : -jHHf, H and 
IS in amperes = i, 1, and 1 amperes. The heats devel- 
oped are, by the rule, in the ratio 100 X } : 35 X 1 
and 50 X 1 or 50: 35 : 50. 

The same problem can be done directly by Eule 
31, thus: The three lamps develop heat in the 
ratio m* : H* : if = 50 : 35 : 50. This is the direct 
and preferable method of calculation. 

Note, — For " heat, ,v " rate of energy," or " rate of work " 
can be read in these rules. 



54 ARITHMETIC OF ELECTRICITY. 

The Joule or Gram-Calorie. 

The last rules and problems only touch upon the 
relative proportions of heat; they do not give any 
actual quantity. If we can express in units of the 
same class a standard quantity of heat, then by deter- 
mining the relation of the standard to any other 
quantity, we arrive at a real quantity. Such a stan- 
dard is the joule, sometimes called the "calorie" or 
"gram-calorie." A joule is the quantity of heat 
required to raise the temperature of 1 gram of 
water 1 degree centigrade. It is often expressed 
as a water-gram-degree C. or w. g. d. C. or for 
shortness g. d. C, from the initials of the factors. 
It is unfortunate that it is called the calorie 
as the name is common to the water-kilogram- 
degree 0. or kg.d. 0. The joule is equal to 4.16 
X 10 7 or 41,600,000 ergs. 

It will be remembered that practical electric units 
are based on multiples of the 0. G-. S. units of which 
the erg is one. The joule comes in the C. Gr. S. 
order. Therefore to determine quantities of heat 
the following is a general rule when the practical 
electric units are used. 

Rule 33. Obtain the expression of rate of heat devel- 
oped, or of rate of energy, or of rate of work in volt 
amperes, Reduce to €. G. S. units (ergs) by multiply- 
ing by 10 7 and divide by the value of a joule in ergs 
(4.16 X 10 7 ). The quotient is joules or water-gram de- 
grees €• per second. 

_ ExCXIO 7 
**~ 4.16 X 10 7 



WORK AND ENERGY. 65 

Example. 

A current of 20 amperes is flowing through a wire. 
What heat is developed in a section of the wire 
whose ends differ in potential by 110 volts? 

Solution: The rate of energy in watts or volt- 
amperes = 110 X 20 = 2200. In the C. G. S. units 
this is expressed by (110 X 10 8 ) X (20 X 10 1 ) = 2200 
X 10 7 ergs, per second; . \ quantity of heat = 2200 X 
10 7 -^ 4.16 X 10 7 = 528.8 joules or gram-degrees- 
centigrade per second. 

As 10 7 -5- 10 7 = 1 the rule can be more simply 
stated thus: 

Rule 34. The quantity of heat produced per second 
in a circuit by a current is equal to the product of the 

watts by — — or by .24. 

Q, = 0.24CE. or 0.24 «5 

Examples. 

A difference of potential of 5.5 volts is main- 
tained at the terminals of a wire of fa ohm resist- 
ance. How many joules per second are developed? 

Solution: By Ohm's law, current = 5.5 -*- fa = 55 
amperes. By the rule 55 X 5.5 X 0.24 = 72.6 
joules per second. 

Note. — The energy of a current is given by Kules 30, 31 
and 32 in watts, so that all cases are provided for by a com- 
bination of one or the other of these rules with Rule 34 
An example will be given for each case. 



; 

I 

N 



WORK 

reduiituuit, while 

■ 

jwer un 

Heat. 

tion. 

- 

KuJt 1 • 

I tilt? lor- - 

jhm. 
tation. (Dicy*) 



58 ARITHMETIC OF ELECTRICITY. 

Solution: By Eules 30 and 34, we find rate of heat 
= .75 2 X .47 = .264375 watts; .264375 X .24 = 
.06345 joules per second. The current is to last for 
300 seconds .'. total joules = .06345 X 300 = 19.035 
joules. This must be divided by the specific heat 
of mercury to get the corresponding weight of mer- 
cury; 19.035 + .0332 = 573 gram degrees of mercury. 
Dividing this by the weight of mercury, 20.25 grams, 
we have 573 + 20.25 = 28° 0. ^ 

Rule 36. By radiation and convection, 4000 Joule 
about is lost by any unpolislied substance in the air for 
each square centimeter of surface, and for each degree 
that it is heated above the atmosphere. 

Example. 

A conductor of resistance, 8 ohms/ has a current 
of 10 amperes passing through it. It is 1 centi- 
meter in circumference, and 100 meters long. How 
hot will it get in the air? 

Solution: By Eule 30, etc., the heat developed per 
second in joules is 10 2 X 8 -*- 4.16 = 192.3 joules. The 
surface of the conductor in centimeters is 10,000 X 
1 = 10,000 sq. cent. It therefore develops heat at 
the rate of 192.3 X 10"* = .01923 joules per second 
per square centimeter of surface. When the loss by 
radiation and convection is equal to this, it will re- 
main constant in temperature. Therefore .01923 
■*■ 4tjW = 76.92, the number of degrees 0. above the 
air to which the conductor could be heated by such 
a current. 



WORK AND ENERGY. 59 

Results of this character are only approximate, 
as the coefficient, *t>W, is not at all accurate. 

Rule 37. Tlie cube of the diameter in centimeters of 
a wire of any given material that will attain a given 
temperature centigrade under a given current is equal 
to tlie product of tlie square of tlie current in amperes x 
Sp. Resistance in microhms of the substance of the 
wire, multiplied by .000391, and divided by the tem- 
perature in degrees centigrade. 

d3 a C a X Sp. Kes. x .000391 
t° 

Examples. 

A lead safety catch is to be made for a current of 
7.2 amperes. Its melting point is 335° 0., and its 
specific resistance 19.85 microhms per cubic centi- 
' meter. What should its diameter be? (Day.) 

Solution: By the rule, the cube of the diameter = 
7.2 2 X 19.85 X .00039 + 335 = .001198. The cube 
root of this gives the diameter in centimeters. It is 
.10582 or .106 cm. 

A copper wire is to act as safety catch for 500 am- 
peres: melting point 1050° C— Sp. Eesistance 1.652 
microhms. What should its diameter be? (Day.) 

_ , ,, „ -i « t x 5002 x 1.652 x .000391 

Solution: Cube of diameter = j^so 

15^418 _ >15 23 # The cube root of this is .5341 centi- 

1050 

meter, the thickness of the wire sought for. 

It will be observed that in this rule no attention 
is paid to the length of the wire, as the effect of a 
current in melting a wire has no reference to its 



60 ARITHMETIC OF ELECTRICITY. 

length. The time of fusion will vary with the spe- 
cific heat, but will, of course, be only momentary, 

Work of a Current. 

Rule 38. The work of a current in a giren circuit is 
proportional to the volt amperes, W =• EC 

Example. 

A current A of 3.5 amperes with difference of 
potential in the circuit of 20 volts is to be com- 
pared to B, a 3 ampere current with a difference of 
potential of 1000 volts in the circuit; what is the 
ratio of work produced in a unit of time? 

Solution: Work of A : work of B :: 3.5 X 20 : 
3 X 1000 or as 70 : 3000 or as 1 : 42-^. 

Rule 39. The work of a current in a given circuit is 
equal to the volt -coulombs divided by the acceleration 
of gravitation (9.81 meters). This gives the result in 
kilogram-meters. (7.23 foot pounds.) T?p ._ E. C.t 

9.81 

Example. 

A current of 20 amperes is maintained in a circuit 
by an E. M. F. of 20 volts. What work does it do 
in one minute and a half (90 sec.)? 

Solution : Work — 20 x 20 x 90 sec -r 9.81 = 3670 
kgmts. and 3670 X 7.23 = 26,534 foot pounds. 

Note. — This is easily reduced to horse-power. 26,534 foot 
pounds in 90 sec. = 17,688 foot pounds in 1 min. 1 H. P. 
= 33,000 foot pounds per min. .\ £$£$$ = .536 H..P. of 
above current and circuit. The same result can be ob- 
tained by Kule41 thus: 2 -^-° = .536 H. P, 



WORK AND EXER GY. 61 

Rule 40. To determine work done by a current in a 
given circuit apply Rules 30, 31 or 32 as the case re- 
quires. These give directly tlie watts, multiply by sec- 
onds and divide by 9.81. The result is kilograni-nie- 
ters. 

Examples. 

10 amperes are maintained for 55 sec. through 15 
ohms. What is the work done? 
Solution : By rule 32, watts = 10 2 X 15 = 1500. 

Work = 1500 X 55 - 9.81 = 8409 kgmts. 

1000 volts are maintained between terminals of a 
lead of 20 ohms resistance. Calculate the work 
done per hour. 

Solution: By Bule 32 watts = 1000 2 - 20 = 50,000. 
One hour = 3G00 sec. Work = 50,000 X 3600 + 9.81 
= 18,348,623 kgmts. 

These rules give the basis for determining the ex- 
pense of maintaining a current. The expense of 
maintaining a horse-power or other unit of power or 
work being known the cost of electric power is at 
once calculable. 

Electrical Horse-Power. 

Power is the rate of doing work or of expending 
energy. In an electric circuit one horse-power is 
equal to such a product of the current in amperes, 
by the E. M. F. in volts as will be equal to 746. 

Rule 41 . The electric horse power is found by multi- 
plying the total amperes of current by the volts or E. M. 
]F. of a circuit or given part of one and dividing by 746. 

H.P- - 746 



62 ARITHMETIC OF ELECTRICITY. 

Example. 

250 incandescent lamps are in parallel or on mul- 
tiple arc circuit. Each one is rated at 110 volts 
and 220 ohms. What electric H. P. is expended on 
their lighting? 

Solution: The resistance of all the lamps in par- 
allel is equal to if? ohm. The current is equal to 
110 + m = 125 amperes. H. P. = 125 X 110 + 746 
= 18iV H. P. or 13& lamps to the electrical H. P. 

As it is a matter of indifference as regards absorp- 
tion of energy how the lamps are arranged, a simpler 
rule is the following, where horse-power required for 
a number of lamps or other identical appliances is 
required. 

Rule 42. multiply together the voltage and amperage 
of a single lamp or appliance ; multiply the product by 
the number of lamps or appliances and divide by 746* 

Example. 

Take data of last problem and solve it. 

Solution: Current of a single lamp = ii$ = *4 am- 
pere. H. P. = 110 X y 2 X 250 + 746 = 18 A H. P. 

When the voltage and amperage are not given 
directly, the missing one can always be calculated 
by Ohm's law and the above rules can then be ap- 
plied. The same can be done by applying following: 

Rnle 43. To determine the electrical horse-power 
applyRules 30, 31, or 32; these give directly the watts; 
multiplying the result by -j- G r dividing by 746 gives 
the horse-power. ' 



WORK AND ENERGY. 63 

Examples. 

A current of 10 amperes is maintained through 50 
ohms resistance. What is the electrical horse- 
power? 

Solution: By rules 30 and 43 we have watts = 
10 2 X 50 = 5000 and electrical horse-power = 5000 
-r 746 or 6.7 H. P. 

An electromotive force of 1500 volts is maintained 
in a circuit of 200 ohms resistance. What is the 
electrical horse-power? 

Solution: By Kules 31 and 43 we have watts = 
1500 2 ■*- 200 = 11,250. Electrical horse-power = 
11,250 + 746 or 15 H. P. (nearly). 

Thus the volt-amperes or watts are units of rate 
of heat energy or of rate of mechanical energy. 
The ratio of joules per second to a horse-power is 
746: 4.16 or 179.3 joules per second = 1 H. P. 
Other ratios of power and heat units will be found 
in the tables. 

Duty and Efficiency of Electric Generators. 

Rule 44, The duty of an electric generator is the quo- 
tient obtained by dividing the total electric energy by 
the mechanical energy expended in turning the arma- 
ture. 



— _ e. H. P. (total) 
u " m. H.P. 



Examples. 



A dynamo is driven by the expenditure of 58 
H. P. Its internal resistance is 10.7 ohm. TH* 



64 ABITHMETIC OF ELECTRICITY. 

resistance of the outer circuit is 150 ohms and it 
maintains a current of 16 amperes. What is its 
duty? 

Solution: The total electrical H. P. is found by 
Eules 30 and 43 to be 16 2 X 160.7 -*■ 746 = 55.1 
H. P. Duty = 55.1 + 58.0 = 951 

The result must always be less than unity; if it 
exceeded unity it would prove that there had been 
an error in some of the determinations. 

Rule 45. The commercial efficiency of a generator is 
the quotient obtained by dividing the electric energy in 
the outer circuit by the mechanical energy expended in 
turning the armature* 

,i -ci-pp e. H. P. (outer circuit) 
C - Eff - = m.H.P. 

Examples. 

What is the commercial efficiency of the dynamo 
just cited? 

Solution: The electrical H. P. of the outer circuit 
is found by the same rules to be 16 2 X 150 -*-746 = 
51.5 commercial efficiency = 51.5 ■*■ 58.0 = 88. 8#. 

Rule 46. The resistance of the outer circuit is to the 
total resistance, as the commercial efficiency is to the 
duty. 

Examples. 

Take the case of the generator last given and 
from its duty calculate the commercial efficiency. 

Solution: 150: 160.7 :: x : 95.0.% x = 88.8 or 
88.8*. 



CHAPTER VIII. 
batteries. 

General Calculations of Current. 

A battery is rated by the resistance and electro- 
motive force of a single cell, which factors are 
termed the cell constants. In the case of storage 
batteries, whose susceptibility to polarization is very 
Blight, the resistance is often assumed to be neglible. 
It is not so, and in practice is always knowingly or 
otherwise allowed for. 

From the cell constants its energy-constant may 
be calculated by Rule 31, as equal to the square of 
its electro-motive force divided by its resistance. 
This expresses its energy in watts through a circuit 
of no resistance. 

There are two resistances ordinarily to be consid- 
ered, the resistance of the battery which is desig- 
nated by R or by n R if the number of cells is to be 
implied and the resistance of the external circuit 
which is designated by r. 

Rule 47. The current given by a battery is equal to 
its electro-motive force divided by the sum of the exter- 
nal and internal resistances. 

o= E 



B-f r 



66 



ARITHMETIC OF ELECTRICITY. 









Six cells in parallel. 




Six cells in series. 




Six cells— two in parallel, three in series. 





®H@) 





Six cells— three in parallel, two in series. 



Arrangement op Battery Cells. 



BATTERIES. 67 

Example. 

A battery of 50 cells arranged to give 75 volts 
E. M. F. with an internal resistance of 100 ohms 
sends a current through a conductor of 122 ohms 
resistance. What is the strength of the current? 

Solution: Current = 75 +■ (100 + 122) = .338 
ampere. This rule has already been alluded to 
under Ohm's law (page 14). 

Arrangement of Cells in Battery. 

In practice the cells of a battery are arranged in 
one of three ways, a: All may be in series; i: all 
may be in parallel; c: some may be in series and some 
in parallel, so as to represent a rectangle, s cells in 
series by p cells in parallel, the total number of cells 
being equal to the product of s and p. 

Other arrangements are possible. Thus the cells 
may represent a triangle, beginning with one cell, 
followed by two in parallel and these by three in 
parallel and so on. This and similar types of arrange- 
ment are very unusual and little or nothing is to be 
gained by them. 

Rule 48. The electromotive force of a battery Is 
equal to the E. in. F. of a single cell multiplied by the 
number of cells in series. 

Rule 49. The resistance of a battery is equal to the 
number of its cells in series, multiplied by the resist- 
ance of a single cell and divided by the number of it* 
cells in parallel. 

-d battery s R 

XV. = 

S 



68 ARITHMETIC OF ELECTRICITY. 

Examples. 

A battery of 50 gravity cells 1 volt, 3 ohms each 
is arranged 10 in parallel and 5 in series. What is 
its resistance and electromotive force? 

Solution: Eesistance = 5 X 3 -*- 10 = 1.5 ohms. 
E. M. F. = 5 X 1 = 5 volts. 

The same battery is arranged all in parallel; what 
is its resistance and E. M. F. ? 

Solution: This gives one cell in series. 

Resistance = 1 X 3 ■*- 50 = .06 ohms. 

E. M. F. =1X1 = 1 volt. 

The same battery is arranged all in series; what is 
its resistance? 

Solution: This gives one cell in parallel. 

Eesistance = 50x3 = 150 ohms. 

E. M. F. = 50 X 1 = 50 volts. 

The current given by a battery is obtained from 
these rules and from Ohm's law. 

Example. 

150 cells of a battery (cell constants 1.9 volts, 
J ohm) are arranged 10 in series and 15 in parallel. 
They are connected to a circuit of 1.7 ohms resist- 
ance. What is the current? 

Solution: The resistance of the battery = ^-^ = 
.333 ohms. The E. M. F. = 10 X 1.9 = 19 volts. 
Current = 19-8- (.333 + 1.7) = 9.34 amperes. 



BATTERIES. 69 

Cells Eequired for a Given Current. 

To calculate the cells required to produce a given 
current through a given resistance and the arrange- 
ment of the cells proceed as follows. 

Rule 50. Calculate the cell current through zero exter- 
nal resistance. Case A. If it is twice as great or more 
than twice as great as the current required apply gRule 

51. Case IB. If less than twice as great and more than 
equal or less than equal and more than one half as 
great as the current required and so on apply Rule 

52. Case C. If the cell current is equal to or is a unit* 
ary fraction (!, 1, A, etc.) of the current required apply 
Rule 53. 4 5 8 

Rule 51. Case A. Divide the required difference of po- 
tential of the outer circuit by the voltage of a single cell 
diminished by the product of the required current mul- 
tiplied by the resistance of a single cell. Arrange the 
cells in series. 

a E-ca 

Examples. 

Five lamps in parallel, each of 100 volts 200 ohms, 
are to be supplied by a battery whose cell constants 
are 2 volts i ohm. How many cells and what 
arrangement are required? 

Solution: €ell current = | = 10 amperes. The 
resistance of the five lamps in parallel (Rule 12) = 
2|o — 4Q } ims# The required current therefore =* 
W = 24 amperes. As 10 exceeds 2* X 2 (Rule 50) it 
falls under case A. By Rule 51 the number of cells 



100 _ __ ioo 

(2JXJ) I 

be divided. The cells must be in series. 



is 2-(£xi) == X ^ 66,6 or 67 cells ' as a cel1 cannot 



70 ARITHMETIC OF ELECTRICITY. 

Proof: The E. M. F. of the 67 cells in series 
= 67 X 2 = 134 volts; their resistance = 67 X | = 
13.4 ohms. The resistance of the lamps in par- 
allel is 40 ohms. Hence by Ohm's law the cur- 
rent = 40 !^ 3 4 ' = 2.51 amperes, the current required. 

The same lamps are placed in series. Calculate 
the cells of the same battery required. Cell current 
= 10 amperes. Current required = ^*jj or | am- 
pere. As 10 exceeds }X2 (Rule 50) it falls again 
under case A. By Eule 51 cells required = g ™ . } 
= 263. 

Proof: Current -= - 52 6 ^f 10 oo = i ampere the current 
required. 526 is the number of cells multiplied by 
the voltage of one cell; 52.6 is the number of cells 
multiplied by the resistance in ohms of a single cell; 
1000 is the resistance of a single lamp, 200 ohms, 
multiplied by the number, 5, of lamps in series. 

Whenever the arrangement and number of cells of 
a battery has been calculated the calculation should 
be proved as above. 

Rale 52. Case B. Group two or more cells in parallel 
so as to obtain by calculation from them through no ex- 
ternal resistance a current twice as great or more than 
twice as great as the required current. Then treating 
the group as if it was a single cell apply Rule 51 to de- 
termine the number of groups in series. 

Example. 
Assume the same lamps in parallel, requiring the 



BATTERIES. 71 

current already calculated of 2i amperes. Assume 
a battery of constants 1 volt .25 ohm, giving a 
cell current of 4 amperes. This is less than 21 X 2 
and more than 24 X lj therefore it falls under Case 

B. 

Solution: A group of two cells in parallel gives 
. T ^ = 8 amperes. 8 exceeds 2J X 2 . \ applying Rule 
49 we have number of groups - 100 ■*- [1 — (2j X 
.125)] = 146 groups in series. Total number of cells 
= 2 in parallel, 146 in series = 292 cells. 

Proof: Current = 146 -*- (40 + 18.25) =2.5 am- 
peres. 

Rule 53* Case €• Place as many cells in series as will 
give twice the required voltage. Place as many cells in 
parallel as will give a resistance equal to that of the 
external circuit. 

Example. 

Assume the same lamps in parallel. Assume a 
battery of cell constants, 1 volt, 4 ohms. The lamp 
current is 21 amperes. The cell current is J ampere. 
The cell current therefore equals (I - 2i) A of the 
required current. This falls under Case C. and is 
solved by Rule 53. 

Solution: Voltage required 100. By the rule 
cells in series = 100 X 2 = 200. These have a re- 
sistance of 800 ohms. To reduce this to the resist- 
ance of the outer circuit, viz., 40 ohms, 800 ■*■ 40 
= 20 cells must be placed in parallel. Total cells =~ 
20 x 200 = 4000 cells. 



72 



ARITHMETIC OF ELECTRICITY. 



Proof: Current = 200 + (40 + 40) = 2.5. 

Rule 54. All cases coming under Case C. may be sim- 
ply solved for the total number of cells by dividing the 
external energy by the cell energy and multiplying by 4. 
This gives the number of cells. 

Example. 

Take as cell constants . 75 volt l| ohm giving \ am- 
pere. Assume 20 lamps, each 50 volts, 50 ohms and 
1 ampere. As i -*- 1 is a unitary fraction (J) Case 
0. applies. 

Solution: Cell energy = \ X .75 = .375 watts. Ex- 
ternal energy = 50 x 1 X 20 = 1000 watts. (1000 + 
.375) X 4-10,666 cells. 

Solution by Eule 53: Voltage required taking 
lamps in series = 20 x 50 = 1000. To give twice 
this voltage requires 2000 -*• .75 = 2667 cells in 
series whose resistance is 2667 X .1.5 = 4000 ohms. 
To reduce this to 1000 ohms we need 4 such series 
of cells in parallel giving 10,668 cells. 

Proof: Current = (J ^ 7 1 ^ 100Q = 1 ampere. 

Slight discrepancies will be noticed in the current 
strength given by different calculations. This is 
unavoidable as a cell cannot be fractioned or di- 
vided. 

Efficiency of Batteries. 

Rule 55. The efficiency of a battery is expressed by di- 
viding the resistance of the external circuit by the total 
resistance of the circuit. 

Efficiency = ^—. — 
±v -p r 



BATTERIES. 73 

Example. 

A battery consists of 67 cells in series of constants 
2 volts i ohm. It supplies 5 lamps in parallel, each 
100 volts 200 ohms constants. What is its effi- 
ciency? 

Solution: The resistance of the battery is 67 X \ 
— 13.4 ohms. The resistance of the lamps is (Rule 
12) H* = 40 ohms. Therefore the efficiency of the 
battery is 40.0 -*■ (40 + 13.4) = .749 or 74.9*. 

Rale 5 6. To calculate the number of battery cells and 
ttieir arrangement for a given efficiency : Express the 
efficiency as a decimal, multiply the resistance of the 
external circuit by the complement of the efficiency 
(1— efficiency) and divide the product by the efficiency; 
this gives the resistance of the battery. Add the two 
resistances and multiply their sum by the current to be 
maintained for the E. M, F. of the battery. Arrange the 
cells accordingly as near as possible to these require- 
ments. 

Examples. 

Five lamps, each 100 volts 200 ohms in parallel 
are to be supplied by a battery of cell constants 2 
volts .4 ohm. The efficiency of the battery is to be 
as nearly as possible 75$. Calculate the number of 
cells and their arrangement. 

Solution: The constants of the external circuit are 
40 ohms (Eule 12) and 100 volts. Applying the 
rule we have [40 X (1— .75)] - .75 - 1( H* = 13* 
ohms, the resistance of the battery. By Ohm's law 
the E. M. F. of the battery = (40 + 13*) X 2.5 = 



74 Arithmetic of electricity. 

133i volts. These constants, 13i ohms and 133i 
volts, require 67 cells in series and 2 in parallel. 

Proof: a. Of efficiency, by Rule 55, -~^i = - 75 
or 75$. b. Of number of cells and of their arrange- 
ment 67 X 2 = 134 volts; (67 X .4) -*- 2 = 13.4 
ohms; 134 + (13.4 + 40) = 2.5 amperes. 

Rule 57. Where a fractional or mixed number of cell* 
In parallel are called for to produce a given efficiency, 
take a group of the next highest integral number of 
eells in parallel and proceed as in Rule 51* 



Example. 

Assume a current of 3i amperes to be supplied 
through a resistance of 30 ohms, absorbing 100 
volts E. M. P. Let the cell constants of a battery 
to supply this circuit be 2 volts, I ohm. Calculate 
the cells and their arrangement for 80 per cent, 
efficiency. 

Solution: By Rule 56 efficiency = .80 and 
— * go"' = 7i ohms, which is the required resist- 
ance of the battery; 74+ 30 = 37i ohms are the total 
resistance of the circuit. By Ohm's law, 37i X 3i = 
125 volts, the required E. M. F. of the battery. 
This requires 63 cells in series, with a resistance for 
one series of 63 X J = 10J ohms. To reduce this to 
7j ohms ~ = 1.4 cells in parallel are required. As 
this is a mixed number we take the next highest in- 
tegral number and place 2 cells in parallel. The 
constants of this group of 2 cells are 2 volts, & 



BATTERIES. 75 

ohm. Applying Rule 51 we have for the number of 
such groups in series; 2 _™ x A) = 58 groups in series. 
As there are 2 cells in parallel the total cells are 116, 
of resistance, 58 X ^ = 4.83 ohms, and of E. M. F., 
58 X 2 = 116 volts. 

Proof: Of efficiency by Rule 55, ^^ = 86.1*. 
Of number and arrangement of cells ^ " 4 m = 3.33 
amperes. 

It is to be observed that the efficiency thus 
obtained is far from what is required. In most 
cases accuracy can only be attained by arranging 
the battery irregularly, which is unusual in prac- 
tice. An example will be found in a later chapter. 

Chemistry of Batteries. 

One coulomb of electricity will set free .010384 
milligrams of hydrogen. The corresponding weights 
of other elements or compounds are found by multi- 
plying this factor by the chemical equivalent, and 
dividing by the valency of the element or metal of 
the base of the compound in question. 

An element or other substance in entering into 
any chemical combination develops more or less 
heat, always the same for the same weight and com- 
bination. The atomic weight of an element or the 
molecular weight of a compound divided by the val- 
ency of the element or metal of its base gives the 
original chemical equivalent. 



76 ARITHMETIC OF ELECTRICITY. 

The quantities of heat evolved by the combination 
of quantities of substances expressed by their original 
chemical equivalents multiplied by one gram are 
termed the thermo-electric equivalents of the ele- 
ments or substances in question. In the tables it is 
expressed in kilogram degrees 0. of water (kilogram- 
calories). 

From the thermo-electric equivalent of a combi- 
nation we find the volts evolved by it or absorbed by 
the reciprocal action of decomposition. 

Rule 58. The volts evolved by any chemical com- 
bination or required for any chemical decomposition 
are equal to the thermo-electric equivalent in kilo- 
gram-calories multiplied by .043. 

E = .043 X H. 

Examples. 

What number of volts is required to decompose 
water ? 

Solution: From the table we find that the com- 
bination of one gram of hydrogen with eight grams 
of oxygen liberates 34.5 calories. Then 34.5 X 
.043 = 1.48 volts. 

Rule 59, To determine the voltage of a galvanic 
couple subtract the kilogram calories corresponding to 
decompositions in the cell from those corresponding to 
combinations in the cell for effective energy and multi- 
ply by .043 for volts. 

Examples. 

Calculate the voltage of the Smee couple. 
Solution: In this battery zinc combines with 



BATTERIES. 77 

oxygen, giving out 43.2 calories and combines with 
sulphuric acid, giving 11.7 more calories; a total of 
54.9 calories. An equivalent amount of water is at 
the same time decomposed acting as counter-energy 
of 34.5 calories. The effective energy is 54.9 — 
34.5 = 20.4 calories. The voltage = 20.4 X .043 = 
.877 volts. 

Calculate the voltage of the sulphate of copper 
battery. 

Solution: Here we have combination of zinc with 
sulphuric acid as above 54.9 calories; decomposition 
of copper sulphate 19.2 + 9.2 = 28.4 . \ 54.9 — 28.4 
= 26.5 calories effective energy 26.5 X .034 = 1.14 
volts. 

It will be noticed that these results are approxi- 
mate. Some combinations are omitted in them as 
either of unknown energy, or of little importance. 

Work of Batteries. 

The rate of work of a battery is proportional to 
the current multiplied by the electro-motive force. 
The work is distributed between the battery and the 
external circuit in the ratio of their resistances as by 
Rule 55. The horse-power, and heating power are 
calculated by Rules 30-43, care being taken to dis- 
tribute the energy acording to the resistance by 
the following rule: 

Rule 60. The effective rate of work or the rate of 
work in the external circuit of a battery, is equal to the 



78 ARITHMETIC OF ELECTRICITY. 

total rate multiplied by tlie efficiency of tlie battery ex- 
pressed decimally* 

Example. 

25 cells of 2 volt 1 ohm battery are arranged in 
series on an external circuit of 250 ohms resistance. 
What work do they do in that circuit? 

Solution: The current (Ohm's law) = J^^ = § 
1.818 amperes. Total rate of work = 1.818 X 50 
volts = 90.9 watts. Efficiency of battery = Iff = 90 
per cent, (nearly). Effective rate of work = 1.818 X 
50 X .90 = 81.81 watts. 

Chemicals Consumed in a Battery. 

Rule 61. The chemicals consumed in grams by a bat- 
tery for one kilogram-meter (7.23 foot lbs.) of work are 
found by multiplying the combining equivalent of the 
chemical by the number of equivalents in the reaction 
by the constant .000101867 and dividing by the pro- 
duct of the E. ilE. F. by the valency of the element in 
question* 

w _ Equiv. X n X ,000101867 
E x valency 

Examples. 

What is the consumption of zinc and sulphate of 
copper per kilogram-meter of work in a Daniel's 
battery? 

Solution: Take the E. M. F. as 1.07 volt. The 
equivalent of zinc, a dyad, is 65 and one atom en- 
ters into the reaction. The zinc consumed there- 
f 0re = 65xl 1 ^2 01867 = -00309 gram. 



BATTERIES. 79 

The equivalent of copper sulphate, is 159.4. One 
equivalent enters into the reaction carrying with it 
one atom of the dyad metal copper. The weight 
consumed therefore - ^Axixmrnm = Qm 

l.U/ x * 

grams. Add 56.46$ for water of crystallization. 

All these quantities are for one kilogram-meter of 
work (7.23 foot lbs.) which may be more or less 
effective according to circumstances as developed in 
Eules 44, 45, and 60. 

Decomposition of Compounds by the Battery. 

In cases where a compound has to be decomposed 
by a battery two resistances may be opposed to the 
work. One is the ohmic resistance of the solution, 
which is calculated by Eule 16. The other is the 
electromotive force required to decompose the solu- 
tion. This is best treated as a counter-electromotive 
force. Then from the known data the current rate 
is calculated, and from the electro-chemical equiva- 
lents the quantity of any element deposited by a 
given number of coulombs is determined. 

Rule 62. To calculate the metal or other element lib- 
erated by a given current per given time proceed as fol- 
lows: Calculate the resistance. Determine the counter- 
electromotive force of the solution by Rule 58 and sub- 
tract it from the E. HI. F. of the battery or generator. 
Apply Ohm's law to the effective voltage thus deter- 
mined and to the calculated resistances to find the cur- 
rent. Multiply the electro-chemical equivalent of the 
element by the coulombs or ampere-seconds. 



80 ARITHMETIC OF ELECTRICITY. 

Example. 

A bath of sulphate of copper is of specific resis- 
tance 4 ohms. The electrodes are supposed to be 
10,000 sq. centimeters in area and 5 centimeters 
apart. Two large Bunsen elements in series of 
1.9 volts .12 ohms each are used. What weight 
in milligrams of copper will be deposited per 
hour? 

Solution: By Kule 16 the resistance of the solu- 
tion is -^~~ = 0.023. The electro-chemical equiva- 
lent of copper is .00033 grams. The thermo-electric 
equivalent for copper from sulphate of copper 
is 19.2 + 9.2 = 28.4 calories. The E. M. F. 
corresponding thereto = 28.4 X .043 = 1.22 volts 
counter E. M. F. The E. M. F. of the bat- 
tery = 1.9 X 2 = 3.8 volts, giving an effective E. M. 
F. of 3.8 — 1.22 = 2.58 volts. The resistance of the 
battery = .12 X 2 = .24 ohms. The current = 
- 9.8 amperes. This gives per hour 9.8 X 



.24 +.023 

3,600 = 35,280 coulombs, and for copper deposited 
.00033 X 35,280 = 11.64 grams. 

In many cases one electrode is made of the mater- 
ial to be deposited and being connected to the car- 
bon end of the battery or generator is dissolved as 
fast as the metal is deposited. In such case there 
is no counter electro-motive force to be allowed 
for. 



BATTERIES. 81 

Example. 

Take the last case and assume one electrode (the 
anode) to be of copper and to dissolve. Calculate 
the deposit. 

Solution: Current = 3.8+ (.24 + .023) = 14.4 
amperes = 51,840 coulombs per hour = .00033 X 
51,840 = 17.10 grams of copper. 



CHAPTER IX. 
electro-magnets, dynamos and motors. 

The Magnetic Field and Lines of Force. 

A current of electricity radiates electro-magnetic 
wave systems, and establishes what is known as a 
field of force. The field is more or less active or in- 
tense according to the force establishing it. The 
intensity of a field is for convenience expressed in 
lines of force. These are the units of magnetic 
intensity, often called units of magnetic flux, and 
the line as a unit is comparable to the ampere X 10, 
which is the C. GL S. unit of current. A line of 
force is that quantity of magnetic flux which passes 
through every square centimeter of normal cross- 
section of a magnetic field of unit intensity. The 
line is at right angles to the plane of normal cross- 
section of such field. Such intensity of field exists 
at the center of curvature of an arc of a circle of ra- 
dius 1 centimeter, and whoge length is 1 centimeter, 
when a current of 10 amperes passes through this 
arc. Practically it is the amount which passes 
through an area of one square centimeter, situated 
in the center of a circle 10 centimeters in diameter, 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 83 

surrounded by a wire through which a current of 
7.9578 amperes is passing. The plane of the circle 
is a cross-sectional plane of the field; a line perpen- 
dicular to such plane gives the direction of the lines 
of force, or of the magnetic flux. 

This cross-sectional area is often spoken of as the 
field of force. As a field exists wherever there are 
lines of force, there are in each magnetic circuit 
either an infinite number of fields of force, or a 
field of force is a volume and not an area. 

The number of lines of force or of magnetic flux 
per unit cross-sectional area of the magnetic cir- 
cuit, i. e. per unit area of magnetic field, expresses 
the intensity of the field. In soft iron, it may run as 
high as 20,000 or more lines per square centimeter 
of cross-section of the iron which is magnetized. 

Just as we might speak of a bar of copper acting 
as conductor for 20,000 C. G. S. units of current, or 
2000 amperes, so we may speak of the iron core of a 
magnet carrying 20,000 lines of force. 

Permeance and Reluctance. 

This action of centralizing in its own material 
lines of force is analogous to "conductance." It is 
termed permeance. Its reciprocal is termed 
reluctance, which is precisely analogous to 
" resistance." Iron, nickel, and cobalt possess high 
permeance; the permeance of air is taken as unity. 
At a low degree of magnetization, soft iron pos- 



84 ARITHMETIC OF ELECTRICITY. 

sesses 10,000 times the permeance of air. At high 
degrees of magnetization, it possesses much less in 
comparison with air, whose permeance is unchanged 
under all conditions. 

There is no substance of infinitely high reluctance, 
which is the same as saying that there is no insula- 
tor of magnetism. 

Magnetizing Force and the Magnetic Cir- 
cuit. 

The producing cause of the magnetic flux or mag- 
netization just described is in practice always a cur- 
rent circulating around an iron core. The name of 
magnetizing force is often given to it. It is the 
analogue of electro-motive force, and is measured by 
the lines of force it establishes in a field of air of 
standard area. 

A high value for the magnetic force is 585 lines 
per square centimeter. It is proportional to the 
amperes of current and to the number of turns the 
conductor makes around it. Its intensity is often 
given in ampere-turns. 

Magnetization always implies a circuit. As far as 
known, magnetic lines of force cannot exist without 
a return circuit, exactly like electric currents. But 
owing to the imperfect reluctance of all materials, 
the lines of force can complete their circuit through 
any substance. In a bar magnet the return branch 
of the circuit is through air. 



ELECT 110-MAGNETS, DYNAMOS AND MOTORS. 8d 

111 the same magnetic circuit, the planes of nor- 
mal cross-section lie at various angles with each 
other. 

The law of a magnetic circuit is exactly compar- 
able to Ohm's law. It is as follows: 

Rule 63. The magnetization expressed in lines of 
force is equal to the magnetizing force divided by the 
reluctance or multiplied by the permeance of the entire 
circuit. 

This rule would be of very simple application, ex- 
cept for the fact that reluctance increases, or per- 
meance decreases, with the magnetization, and the 
rate of variation is different for different kinds of 
iron. 

Rule 64. Permeability is the ratio of magnetization 
to magnetizing force, and is obtained by dividing mag- 
netization by magnetizing force. 

Permeability has to be determined experimentally 
for each kind of iron. It is simply the expression 
of a ratio of two systems of lines of force. It 
always exceeds unity for iron, nickel, and cobalt. 
The specific susceptibility of any particular iron to 
magnetization is its permeability. The susceptibil- 
ity of a portion, or of the whole of a magnetic cir- 
cuit is its permeance. 

General Kules for Electro-Magnets. 

The traction of a magnet is the weight it can 
sustain when attached to its armature. It is pro- 



86 ARITHMETIC OF ELECTRICITY. 

portional to the square of the number of lines of 
force passing through the area of contact. 

Rule 65. The traction of a magnet in pounds is equal 
to the square of the number of lines of force per square 
inch, multiplied by the area of contact and divided by 
7 2,13-1,000. In centimeter measurement the traction 
in pounds is equal to the square of the number of lines 
of force per square centimeter multiplied by the area of 
contact and divided by 11,183,000. The traction in 
grams is equal to the latter dividend divided by 24,655 
(8 * x 981); for dynes of traction the divisor is 25.132 
(8w) 

Examples. 

A bar of iron is magnetized to 12,900 lines per 
square inch; its cross-section is 3 square inches. 
What weight can it sustain, assuming the armature 
not to change the intensity of magnetization? 

Solution: 12,900 2 X 3 = 499,230,000. This di- 
vided by 72,134,000 gives 6.914 lbs. traction. 

A table calculated by this rule is given. A 
diminished area of contact sometimes increases trac- 
tion, and a non-uniform distribution of lines may 
occasion departures from it. The above rule and 
the table alluded to are practically only accurate 
for uniform conditions. The reciprocal of the rule 
is applied in determining the lines of force of a 
magnet experimentally. 

Rule 66. The lines of force which can pass through a 
magnet core with economy are determined by the tables, 
keeping in mind that it is not advisable to let the per- 
meability fall below 200—300. From them a number 
is taken (40,000 lines per square inch for cast iron or 



ELFA'TRO-MAGNETS, DYNAMOS AND MOTORS. 87 

1 00,000 linos per square inch for wrought Iron are good 
general averages) and is multiplied by the cross-sec- 
tional area of tlie magnet core. 

Rule 67. To calculate the magnetizing force in am- 
pere turns required to force a given number of magnetic 
lines through a given permeance, multiply the desired 
lines of force by the reluctance determined as below. 

Rule 68. a. The reluctance of a core or of any portion 
thereof for Inch measurements is equal to the product 
of the length of the core or of the portion thereof by 
0.3132 divided by the product of its cross-sectional area 
and permeability. 

b. The reluctance for centimeter measurements is 
equal to the length of the core divided by the product of 
1.2566, by the cross-sectional area and the permea- 
bility. 

Examples. 

440,000 lines are to be forced through a bar of 
wrought iron 10 inches long and 4 square inches in 
area; calculate its reluctance and the magnetizing 
force in ampere turns required to effect this mag- 
netization. 

Solution: The reluctance (a) = 10 X .3132 + (4X 
permeability). 440,000 lines through 4 square inches 
area is equal to 110,000 lines through 1 square 
inch; for this intensity and for wrought iron the 
permeability = 166. 166 X 4 = 664. The reluc- 
tance therefore = 3.132 ■*- 664 = .0047. The mag- 
netizing force in ampere turns = 440,000 X .0047 
= 2068. 

The same number of lines are to be forced 
through a bar 25.80 square centimeters area and 



88 ARITHMETIC OF ELECTRICITY. 

25.40 centimeters long. Calculate the ampere 
turns. 

Solution: 440,000 lines through 25.80 sq. cent. 
= ~^- = 17,054 through 1 sq. cent., for which the 
permeability = 161. The reluctance therefore, (b) 
= 25.40 -*- (1.2566 X 25.80 X 161) =.0048. The 
ampere turns = 440,000 X .0048 = 2112. 

Magnetic Circuit Calculations. 

Practically useful calculations include always the 
attributes of a full magnetic circuit, because mag- 
netization can no more exist without a circuit than 
can an electric current. In practice an electro- 
magnetic circuit consists of four parts: 1, The 
magnet cores; 2 and 3, the gaps between armature 
and magnet ends; 4, the armature core. To cal- 
culate the relations of magnetizing force to magne- 
tization the sum of the reluctances of these four 
parts has to be found. A further complication 
is introduced by leakage. The permeability of well 
magnetized iron being so low, not exceeding 150 to 
300 times that of air, a quantity of lines leak across 
through the air from magnet limb to magnet limb. 
Leakage is included in the sum of the reluctances 
by multiplying the reluctance of the magnet core by 
the coefficient of leakage, which is calculated for 
each case by more or less complicated methods. For 
parallel cylindrical limb magnets the calculation is 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 89 

exceedingly simple. The calculation in all cases is 
simplified by the fact already stated, that in air 
permeability is always equal to unity, whatever the 
degree of magnetization. For copper and other 
non-magnetizable metals the variation from unity is 
so slight that it may, for practical calculations, be 
treated as unity. 

Leakage of Lines of Force. 

Leakage is the magnetic flux through air from 
surfaces at unequal magnetic potential, such as 
north and south poles of magnets. It is measured 
by lines of force and is proportional to the relative 
permeance of its path. 

The coefficient of leakage of a magnetic circuit is 
the quotient obtained by dividing the total magnetic 
flux by the flux through the armature. The total 
magnetic flux is the maximum flux through the 
magnet core. 

Rule 69. To obtain the coefficient of leakage divide 
the permeance of the armature core and of the two gaps 
pins one-half the permeance of air- between magnet 
limbs by the permeance of the armature core and of the 
two gaps. 

Example. 

The total flux through an armature core is found 
to be at the rate of 70,000 lines per square inch, and 
the armature core is 3 inches diameter and 10 inches 
long. The average length of travel of the magnetic 



90 ABITHMETIC OF ELECTRICITY. 

lines through it is li inches. The air gaps are 10 X 
3 inches area and i inch thick. The permeance be- 
tween the limbs of the magnet is 500. Calculate 
the coefficient of leakage. 

Solution: 70,000 lines per square inch gives a 
permeability of 1,921. By Rule 68 the reluctance 
of the armature core is 30 *\ m X .3132 = .000008. 
The reluctance of a single air gap is ± X .3132 = 
.0052. Thus the armature reluctance is so small 
that it may be neglected. The permeance of the 
two air gaps is given by 0052 1 x 2 = 100 (about). The 

coefficient of leakage = m + 260 = 3.5. 

As the coefficient of leakage is the factor used in 
these calculations, the permeance of the leakage paths 
is the desired factor for its determination. In the 
case of cylindrical magnet cores parallel to each 
other, they are obtained from Table XIII. given in 
its place later. It is thus calculated and used. 
The least distance separating the cores (b) is di- 
vided by the circumference of a core (p) giving the 
ratio (-) of least distance apart to perimeter of a 
core. The number corresponding in columns 3 or 5 
is multiplied by the length of a core. The product 
is the permeance. Columns 2 and 4 give the re- 
luctance. To reduce to average difference of mag' 
netic potential divide by 2. 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 91 

Example. 

Calculate the permeance between the legs of a 
magnet, 3 inches in diameter and 12 inches high 
and 5 inches apart. 

Solution: The perimeter = 3X3.14 = 9.42. ^ 2 = 
i or .5 nearly. From the table of permeability 
we find 6.278. Multiplying this by 12 we have 
6.278 X 12 = 75.336, the permeance. Dividing by 
2 we have '-^|^ = 37.668, the permeance for use in 
the calculation of leakage coefficient. 

It will be observed that this calculation is based 
entirely on the ratio stated, and that absolute di- 
mensions have no effect on it. 

For flat surfaces, parallel and facing each other, 
the following method precisely comparable to the 
rule for specific resistance is used: 

Rule 70. The permeance of the air space between flat 
parallel surfaces is equal to their average area multi- 
plied by 3.193 and divided by their distance apart, all 
in inch measurements. 

Example. 

Determine the permeance between the two facing 
sides of a square cored magnet 15 inches long, 3 
inches wide and 8 inches apart. 

Solution: 3 X 15 = 45 (the average area); 45 X 
3.193 + 8 = 17.96. For use in calculations it 
should be divided by 2 giving 8.98. This division 
by 2 is to reduce it to the average difference of mag- 
netic potential between the two magnet legs. 



92 ARITHMETIC OF ELECTRICITY. 

Calculations for Magnetic Circuits. 

A magnetic circuit is treated like an electric one. 
The permeance (analogue of conductance) or reluc- 
tance (analogue of resistance) is calculated for its 
four parts, magnet core, two air gaps, and armature 
core. The leakage coefficient is determined and ap- 
plied. The requisite magnetizing force is calcu- 
lated in the form of ampere turns (the analogue of 
volts of E. M. F.). The preceding leakage rules 
cover the case of parallel leg magnets. For others 
a slight change is requisite in the leakage calcula- 
tions, but in practice an average can generally be 
estimated. 

Example. 

Assume the magnet and armature of a dynamo. 
The magnet is of cast iron, each leg is cylindrical in 
shape, 4 inches in diameter and 20 inches high. 
From center to center of leg the distance is 9 inches. 
The armature core of soft wrought iron is 4 inches 
in diameter and 8 inches long, the pole pieces curv- 
ing around it are 4 inches, measured on the curve 
inside, by 8 inches long. The air gap is i inch 
thick. Calculate the reluctance of the circuit and 
the ampere turns for 500,000 lines of force. 

Solution: The pole pieces approach within 2i 
inches of each other. This leaves l£ inches of the 
diameter of the armature core embedded or included 
within or embraced by them. One-half of this 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 93 

amount may be taken and added to 2i giving 3J as 
the average depth of core for an area 4 X 8 = 32 
square inches. The lines per square inch of arma- 
ture core are 50 ^ 00 = 15,625 lines per square inch. 
By the table of permeability, 4650 is given for per- 
meability for 30,000 lines in soft iron. For 15,625 
lines per square inch 9,000 can safely be taken for 
permeability. Its relative reluctance is therefore 
32x9000 = '000011 relative armature core reluctance. 

(i) 

The relative reluctance of one air gap (permeabil- 
ity = 1) is i + 32 = .0078 and .0078 X 2 = .0156 =* 
air gaps reluctance (2). 

The ratio - of the table for determining the leak- 



p 



age between cylindrical magnet legs is 4x314 = .4. 
5 is the distance between the legs. Permeance cor- 
responding thereto is 6.897, which multiplied by 20, 
the length of the legs, and divided by 2 for average 
magnetic potential difference gives 68.97 for rela- 
tive effective permeance (3). 

The relative reluctance of the air gaps and arma- 
ture core is .015611; the reciprocal or permeance is 
64.06 (4). 

For coefficient of leakage we have (64.06 + 68.97) 
s- 64.06 = 2.08 (5). 

To find the relative reluctance of the magnet core 
whose yoke may be taken as of mean length 9 inches 



94 ARITHMETIC OF ELECTRICITY. 

and of area equal to that of the core (3.14 X 2 2 = 
12.56) we have to first determine the permeability. 
^m l = 40,000 lines per square inch, corresponding 
to a permeability of 258. For the effective reluc- 
tance of the magnet core introducing the factor of 
leakage (2.08) we have the expression (20 +f 5 + x 9) ^ 8 2 ' 08 
= .0314. 

To get ampere turns, we add the reluctances of 
circuit, multiply by .3132 and by the required lines, 
(.000011 + .0156 + .0314) X .3132 X 500,000 = 
7362 ampere turns required. 

In the above calculations, the multiplication by 
.3132 was omitted to save trouble, relative reluc- 
tances only being calculated, until the end when one 
multiplication by .3132 brought out the ampere 
turns. The leakage appears excessive partly be- 
cause of the high reluctance of the two air gaps. 
These should be increased in area and reduced in 
depth if possible. The leakage is also high on 
account of the legs of the magnet being close to^ 
gether. Were these separated, a larger armature 
core might be used, justifying a lower speed or rota- 
tion of armature, reducing reluctance of air gaps by 
increasing their area, and reducing leakage between 
magnet legs by increasing their distance. The 
magnet legs might also be made shorter, thus reduc- 
ing leakage. 






ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 95 

Thus assume the magnet core of the same cross- 
sectional area, but only 10 inches long and with a 
distance apart of legs of 7 inches, giving a 7 X 10 
inch armature core and pole piece areas (air gap 
areas) of 7 X 10 = 70 sq. inches. 

For leakage ratio we have (^) = j^gg = .56 giv- 
ing from the proper table 6.000 (about), 6 '°°° 2 X 10 
= 30 relative permeance of air space between legs. 

For air gaps reluctance i •*. 70 = .00357 which 
for the two gaps gives .00714 relative reluctance. 

Treating the armature core as a prism 7 X 10 = 70 
sq. inches area and 5 inches altitude, we have for 
lines per sq. inch 500,000 -*- 70 = 7000 giving it 
about 9000 and reluctance as 5 «*- (70 X 9,000) = 
.000008 reluctance. 

Air gaps and armature core reluctance = .007143 

and permeance = -qq^ = 139. 

Coefficient of leakage = 1 -?^p = 1.21. 

If the depth of the air gaps was reduced to $■ inch 
the coefficient of leakage would then be about 1.11. 

Every surface in a magnet leaks to other surfaces 
and the leakage from leg to leg is sometimes but one 
third of the total leakage. In practice the total 
leakage often runs as high as 50$, giving a coefficient 
of 2.00 and in other cases as low as 25#, giving a co- 
efficient of 1.33. 



96 ABITHMETIC OF ELECTRICITY. 

Dynamo Armatures. 

An armature of a dynamo generally comprises two 
parts — the core and the winding. The core is of 
soft iron. Its object is to direct and concentrate 
the lines of force, so that as many as possible of 
them shall be cut by the revolving turns or convolu- 
tions of wire. The winding is usually of wire. It 
is sometimes, however, made of ribbon or bars of 
copper. Iron winding has also been tried, but has 
never obtained in practice. The object of the wind- 
ing is to cut the lines of force, thereby generating 
electro motive force. The number of the lines of 
force thus cut in each revolution of the armature 
is determined from the intensity of the field per 
unit area, and from the position, area and shape of 
the armature, coils and pole pieces. The number 
thus determined, multiplied by the number of times 
a wire cuts them in a second, and by the effective 
number of such wires, gives the basis for determin- 
ing the voltage of the armature. 

Rule 71. One volt E. M. F. Is generated by the cutting 
of 10 8 (100,000,000) lines of force in one second. 

Examples. 

A single convolution of wire is bent into the form 
of a rectangle 7 X 14 inches. It revolves 25 times a 
second in a field of 20,000 lines per square inch. 
What E. M. F. will it develop at its terminals? 

Solution: The area of the rectangle is 7 X 14 = 98 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. ( J7 

square inches. Multiplying this by the lines of force 
in a square inch, we have 98 X 20,000 = 1,960,000. 
Each side of the rectangle cuts these lines twice in 
a revolution, and makes 25 revolutions in a second. 
This gives 25 X 2 X 1,960,000 = 98,000,000 lines 
cut per second, corresponding to 98 X 10 6 X 10~ 8 = 
98 x 10" 2 = r 9 A volts E. M. F. generated, or ?%**%%%%% 
= AV volts. 

The field of the earth in the line of the magnetic 
dip = .5 line per square centimeter. Calculate a 
size, number of layers, and speed of rotation for a 
one volt earth coil. 

Solution : We deduce from the rule the following : 
Area of coil X revolutions per sec. X convolutions 
of wire X .5 X 10~ 8 = .5. "We may start with revolu- 
tions per second, taking them at 20. Next we may 
take 50,000 convolutions. 20 X 50,000 x .5 = 
500,000. This must be multiplied by 200 to give 
10 8 ; in other words, the average area within the 
wire coils must be 100 square centimeters, or 10 X 
10 centimeters. 2 x 100 X 2000 X 500 X .5 = 10 8 , 
and 10 8 X 10" 8 = 1 volt 

Rule 72. Tlie capacity of an armature for current Is 
determined by the cross-section of its conductors. This 
should be such as to allow 520 square mils per ampere 
= 1923 amperes per square inch area. 

Example. 

A drum armature coil is of 4 inches diameter, 

and is wound with wire jh of the periphery of the 



98 ARITHMETIC OF ELECTRICITY. 

drum in diameter; the wire is 100 feet long. Its 
E. M. F. is 90 volts. What is the lowest admissible 
external resistance? 

Solution: The circumference of the drum is 3.14 
X 4 = 12.56 inches. The diameter of the wire is 
^P = .0418 in. or 42 mils. The area of the wire is 
21 2 X 3.14 = 1387 square mils. By the rule the 
allowable current in amperes for a single lead of 
such wire is 1 H3- = 2.66 amperes. But on a drum 
armature the wire lies with two leads in parallel. 
Hence it has double the above capacity or 2.66 X 2 
= 5.32 amperes. The resistance of such wire may 
be taken at .137 ohm. By Ohm's law the total re- 
sistance for the current named must be ^ or W 
ohms. The external resistance is given by 17 ■*- 
.137 = 16.863 ohms. 

These two rules enable us to calculate the ca- 
pacity of any given armature. Certain constants 
depending on the type of armature have to be intro- 
duced in many cases. 

Dkum Type Closed Circuit Armatures. 

For these armatures the following rules of varia- 
tion hold, when they do not differ too much in size, 
and are of identical proportions. 

Rule 73. a. The E. M. F. varies directly with the 
square of the size of core and with the number of turns 
of wire. 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 99 

b. The current capacity varies with the sixth root 
of the wize of core for Identical E. UI. F. 

c. The resistance varies directly with the cube of the 
number of turns and inversely with the size of core. 

d. The amperage on short circuit varies directly with 
the cube of the size and inversely with the square ol 
the number of turns. 

In these rules the proportions of the drum are 
supposed to remain unchanged. Size may be re- 
ferred to any fixed factor such as diameter, as lineal 
size is referred to. 

These rules enable us to calculate an armature 
for any capacity and voltage. As a starting point a 
given intensity of field, speed of rotation, and num- 
ber of turns of wire and size of wire has to be taken. 
The wire is selected to completely fill the periphery 
of the drum. Then a trial armature is calcu- 
lated of the required voltage and its amperage is 
calculated. With this as a basis, by applying Eule 
73, sections a and J, the size of an armature for the 
desired current capacity is calculated, the E. M. F. 
being kept identical. As a standard for medium 
sized machines 20$ of the turns of wire may be con^ 
sidered inactive. 

Example. 

Calculate a 100 volt, 20 ampere armature, whose 
length shall be twice its diameter, to work at a 
speed of 15 revolutions per second. 

Solution: Take as intensity of field 20,000 lines 
per square inch. Allow 80$ of active turns of 



H.ofC. 



100 ARITHMETIC OF ELECTRICITY. 

wire. Start with a core 8X16 = 128 square inches, 
including 128 X 20,000 = 256 X 10 4 lines of force. 
The given speed is 15 rotations per second. For 
the number of active turns of wire per volt we 
have to divide 10 8 or 100,000,000 by one half the 
lines of force cut by one wire per second. This 
number is 256 X 10 4 X 15, or 38,400,000; and 

^XS =2 ' 6 turns - For 10 ° volts > therefore > 260 
active turns are needed. If one half the lines were 
not taken the result would be one half as great, be- 
cause each line cuts each line of force twice in a 
revolution, and in the computation a single cutting 
per revolution only is allowed for. 

The reason for thus taking one half the lines cut 
by a single wire as a base is because in the drum arma- 
ture the wires work in two parallel series, giving one 
half the possible voltage. The actual turns are 
260 + .80 = 325, say 324 turns. Assume it to be 
laid in two layers giving 162 turns to the layer. 
The space occupied by a wire is equal to the peri- 
meter divided by the number of wires or t¥? = .154 
in. Allowing 25$ for thickness of insulation, lost 
space, etc., we have .115 in. or 115 mils as the 
diameter of the wire. In the drum armature as just 
stated the wire is parallel, so that the area of one 
lead of wire has to be doubled, giving 10,573 X 
2 square mils as the area of the two parallel leads. 
This is enough for 40 amperes or double the amper- 
age required. This capacity is reached by taking 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 101 

520 square mils per ampere as the proper cross-sec- 
tional area of the wire. (Rule 72.) 

We must therefore reduce the size to give j4 the 
ampere capacity; this reduction (by Rule 73 b) is in 
the ratio 1 : j£* = 1 : .89018 ; the size therefore is 
8 X .89018 diameter by 16 X .89018 length = 7.12 
X 14.24: inches. 

Applying a for voltage we have for the same num- 
ber of turns on the new armature a voltage in the 
ratio of 1 : .89018* or about A of that required. We 
must therefore divide the number of turns in the 
trial armature by .89018 2 , giving for the number of 
turns |^ = 409, say 410 turns. 

To prove the operation w T e first determine the 
voltage of the new armature. Its area is 7.12 X 
14.24 = 101.4 square inches including 2,028,000 lines 
of force. The active wires are 410 X .8 = 328. 
We have for the voltage = ^y xl5 = 99.78 
volts. 

The relative capacity of the wire is deduced from 
the square of its diameter. The circumference of 
the new armature is 7.12 X 3.14 = 22.3568. There 
are 205 turns in a layer giving as diameter of wire 
^|^ = .1091 mils. This must be squared, giving 
.0119, and compared with the square of the corre- 
sponding number for the original armature. This 
number was 25 -*- .162 = .154 inch. .154 2 = .02371 



102 



ARITHMETIC OF ELECTRICITY. 



and .01190 + .02371 = i (nearly), showing that 
the new armature has one half the ampere capac- 
ity of the old, or 40 X | = 20 amperes as re- 
quired. 

The gauge of the wire is reached by making the 
same allowance for insulation and lost space, viz., 
25fc .1091 X .75 = .0818 in. or 81.8 mils diameter, 
for size of wire. Of course there is nothing absolute 
about 25$ as a loss coefficient; it will vary with 
style of insulation and even to some extent with the 
gauge of wire. But as Rule 73 is based upon the 
assumption that this loss is a constant proportion of 
the diameter of the wire, too great a variation of 
sizes should not be allowed in its application. In 
other words the trial armature should be as near as 
possible in size to the final one. 

Suppose on the other hand that an armature for 
100 amperes was required. This is for 2}4 times 
40 amperes (the capacity of the first calculated or 
trial armature). 

Applying b we extract the 6th root of 2^. (2^)4 
= 1.1653 (by logarithms or by a table of 6th 
roots). The size of the new armature is therefore 
8 X 1.1653 by 16 X 1.1653 or 9.3224 X 18.6448 
inches. 

Applying a for voltage we have for the same num- 
ber of turns of the new armature a voltage in the 
ratio of 1.1653 2 : 1 or 1.358 times too great. We 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 103 

must therefore multiply the original turns by the 
reciprocal of 1.358, giving ^ = 239 turns. 

To prove the voltage, we multiply 239 by .8 for 
the active turns of wire, giving 191.2 turns. The 
area of the armature is 9.32 X 18.64 = 172.7 square 
inches. For voltage this gives 172.7 X 191.2 X 20,- 
000 X 15 X 10" 8 = 98.6 volts (about). 

To prove the capacity we must divide the circum- 
ference of the new armature, 9.32 X 3.14 = 29.26 
inches, by the turns of wire in one layer, H 1 = 120 
turns (about). This gives a diameter of 244 mils 
(nearly). The ratio of capacities of the original and 
this wire is .244* -*- .154 2 inches = .059536 -*- .02371 
=2.51 corresponding to 40 X 2.51 = 100 amperes. 

These results, owing to omissions of decimals, do 
not come out exactly right and it is quite unneces- 
sary that they should. The accuracy is ample for 
all practical purposes. For armature dimensions it 
would be quite unnecessary to work out to the sec- 
ond decimal place. It would answer to take as ar- 
mature sizes in the two cases given 7 X 14^ inches 
and 9# X 18^ inches. 

It is also to be noted that a very low rate of rota- 
tion was taken. 25 to 30 turns per second would 
not have been too much. The latter would give 
double the voltage and the same amperage. 

Field Magnets of Dynamos. 

The calculation for a magnetic circuit given on 



104 ARITHMETIC OF ELECTRICITY. 

pages 92 et seq., is intended to supply an example of 
the calculation of the circuit formed by a field mag- 
net and its armature, such as required for dynamos. 
The leakage of lines of force is and can only be so 
incompletely calculated that it is probably the best 
and most practical plan to assume a fair leakage 
ratio and to make the magnet cores larger than re- 
quired by the lines of force of the armature in this 
ratio. A low multiplier to adopt is 1.25, which is 
lower than obtains in most cases; 1.50 is probably a 
good average. 

Rule 74. The cross-sectional area of the field-magnet 
cores is equal to the lines of force in the field divided by 
the magnetic flux (column B) for the material selected 
and corresponding to the chosen permeability (/a), mul- 
tiplied by the leakage coefficient. 

A good range for permeability is from 200 to 400 
giving for wrought iron from 100,000 to 110,000 
lines of force per square inch and for cast iron from 
35,000 to 45,000 lines per square inch; for the field 
from 15,000 to 20,000 lines per square inch may be 
taken. 

The permeability table gives data for different 
qualities of iron. 

Example. 

Taking the 100 volt 100 ampere armature last cal- 
culated, determine the size of field-magnet cores to 
go with it, and the ampere turns and other data. 

Solution: Assume 20,000 lines of force per square 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 105 

inch in the field, 45,000 in cast iron and 110,000 in 
wrought iron core and a leakage coefficient of 1.25. 
We have for total lines of force passing through 
armature 172.7 X 20,000 = 3,454,000; cross-sectional 
area for cast iron core ^^° X 1.25 = 96 square 
inches; cross-sectional area for wrought iron core 

3,454,000 w i ok on i 

110>000 X 1.^5 = 69 square inches. 

As length of cores we may take 20 inches with 
a distance between them of 10 inches. Assume 
wrought iron to be selected. If cylindrical they 
would be 7 inches in diameter to give the required 
cross-sectional area. The yoke connecting them 
would average in length 10 + 7 = 17 inches, giving 
for magnet cores and yoke a length of 17 + 20 + 20 
= 57 inches. The reluctance of cores and yoke 
(Rule 68) = 5 y 2 x x 8 2 ^ 2 (taking t>> = 200) which reduces 
to .00132 (1). 

The armature area is 172.7 inches. As average 
length of the path of lines of force through it 5 
inches may be taken. As it passes only 20,000 lines 
of force per square inch of field its permeability is 
high, say 9000. Its reluctance is given by ^x^oS * 

This is so low that it may be neglected. 

The area of each air-gap may be taken as 173 
square inches, and of depth of two windings plus 
about A inch for clearance or windage giving 
(.224 X 2) + .1 = about .6 inch for its depth. Its 



106 ARITHMETIC OF ELECTRICITY. 

reluctance is ' 6 x 173 3132 = .00108. As there are two air 
gaps we may at once add their reluctances giving 
.00216 (2). 

By Rule 67 the ampere turns are equal to the 
product of the reluctances (1) and (2), by the lines 
of force giving (.00121 + .00216) X (172.7 + 20,000) 
= 11640 ampere turns. 

The proper size of wire for series winding may be 
determined by Sir William Thompson's rule that in 
series wound dynamos the resistance of the field mag- 
net windings should be j- that of the armature. The 
length of the wire in the armature is equal approxi- 
mately, to the circumference 9.32 X 3.14 = 29.26 
multiplied by the number of turns (240) giving 
29.26 X 240 = 7022 inches. 

The wire turns on the field magnets are found by 
dividing the ampere turns by the amperes giving 
u£p = 232 turns. The circumference of the mag- 
net leg is 7.0 x 3.14 = 22 inches. The total length 
of wire is therefore, approximately, 232 X 22 = 5104 
inches. 

To compare the resistances we must use 1Q p- for 
the length of the armature wire, because it is in 
parallel, and therefore is \ the length and J- the re- 
sistance of the full wire in one length. Dividing by 
4 introduces this factor. 

As the resistances of the wires are to be in the ratio 
of 2 : 3, we have by Rule 13 (calling the thickness 
of armature wire 244 X .75 = 183 mils to allow for 



ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 107 

insulation, etc.), 2:3 :: 183 2 X 5104 : x 2 X *£*, and 
solving we find x 2 = 73026 .'. x = 270 mils. 

For shunt winding Sir William Thompson's rule 
is that the product of armature and field resistance 
should equal the square of the external resistance. 
The latter may be taken (Ohm's law) as equal to 

loo^mperes = 1 ohm - Pro P erl y the armature resist- 
ance should be allowed for, but it is so small that it 
need not be included. We have therefore, arma- 
ture resistance X field resistance = l 2 X 1. 

The armature resistance is .0419 ohms. There- 
fore the field resistance is -^ = 24 ohms. The cur- 
rent through this is equal to h°£- = 4 amperes 
(nearly). Therefore rL ir ± = 2910 turns of wire are 
needed. The length of such wire will be ™*^9 
= 5335 feet. The resistance is about 4. 4 ohms per 
1000 feet corresponding to about .48 mils diameter. 

The Kapp Line. 

Mr. Gisbert Kapp, 0. E. who has given much in- 
vestigation to the problems of the magnetic circuit 
and especially to dynamo construction, is the orig- 
inator of this unit. He considered the regular C. 
G. S. line of force to be inconveniently small. He 
adopted as a line of force the equivalent of 6000 O. 
G. S. lines and as the unit of area one square inch. 
Therefore to reduce Kapp lines to regular lines of 
force they must be multiplied by 6000, and ordi- 



108 ARITHMETIC OF ELECTRICITY. 

nary lines of force must be divided by 6000 to obtain 
Kapp lines. These lines are often used by English 
engineers. The regular system is preferable and by 
notation by powers of ten can be easily used in all 
cases. 



CHAPTER X. 
electric railways. 

Sizes of Feeders. 

To calculate the sizes of feeders for a trolley line 
Rules 23, 24, and 25 in Chapter V. will be found use- 
ful in conjunction with the following ones : 

A. For load at end of feeders: 

Rule 75. The cross-section of tlie feeder in cir- 
cular mils is equal to tlie product of 10.79 times the 
current in amperes times the leugth of the con- 
ductor in feet, divided by the allowable drop in 
volts. 

Example. 

What should be the cross section of a feeder 3,000 
feet long carrying 90 amperes with 35 volts drop? 

- lution: 3,000 X 00 X 10.79 = 2,913,300. Di- 
viding this by 35 gives 83,237 circular mils. This 
would correspond tq a Xo. 1 wire, which has 83,694 
car. mils area. 

All computations of this kind should be checked 
by table XVI. of current capacity on page 153 in 
order to be sure that the wire will not become heated 
above the allowable limit. 



110 ARITHMETIC OF ELECTRICITY. 

Keferring the above example to this table, it is 
seen that a No. 1 wire will carry 80 amperes with a 
rise in temperature of 18° F., and 110 amperes with 
a rise of 36° F. Hence the current of 90 amperes 
will cause a rise of 24° F. This is calculated by- 
simple proportion; subtracting 80 from 90 and from 
110 gives 10 and 30 as the respective differences and 
shows 90 to lie at just one-third the distance from 
80 to 110. Hence the resulting temperature rise 
will be at one-third the distance between 18° and 
36°. The difference between these last two figures 
is 18°, one-third of which is 6°. Add this 6° to 
18° gives us 24° F. as the answer. 

It is often desirable to compute the drop on a 
feeder carrying a given current; this is done by the 
following : 

Rule 76. The drop In volts on any conductor is 
found by multiplying together 10.79, the current in 
amperes and its length in leet, then divide this 
product by its area in circular mils. 

Example. 

What is the drop on a feeder 2,800 feet long, of 
105,592 cir. mils area and carrying a current of 125 
amperes ? 

Solution: 10.79 X 125 X 2,800 = 3,776,500. 

Dividing this product by 105,592 gives 35.76 
volts drop. 

B. For a uniformly distributed load: 



ELECTRIC RAILWAYS. Ill 

The effect of a uniform distribution of load along 
a main or trolley wire is the same as that of half 
the total current passing the full length of the wire; 
hence w^e require but half the cross-section needed to 
deliver the current at the extremity of the wire. 

This is readily done by substituting the constant 
5.4 in place of 10.79 in the foregoing rules. 

In designing electric railway circuits where the 
track forms the path for the return current the rails 
should be of ample area and well bonded, with an 
extra bare wire connected to the bonds and materially 
reducing the drop in the track circuit. 

Power to Move Cars. 

At ordinary speeds on a level track in average con- 
dition it is safe to assume that the force necessary 
to move a car is 30 pounds per ton of weight of car. 

Rule 77. To find the force required to pull or 
push a car on a level track: in average condition, 
multiply the weight of the car in tons by 30* 

Example. 

Find the force required to drag a car weighing 7 
tons on a level track. 

Solution: 7 X 30 = 210 lbs. Ans. 

Should it be required to find the force needed to 
start a car on a level, or to propel it when round- 
ing a curve, substitute the constant 70 in place of 
30 in the foregoing rule. 



112 ARITHMETIC OF ELECTRICITY. 

Example. 

What force is needed to start an 8 ton car on a 
level track? 

Solution: 8 X 70 = 560 pounds. Ans. 

As the above does not take into account the speed 
of the car we shall have to add this factor in order 
to find the horse power needed to move it; we will 
also allow for the efficiency of the motors. 

Rule 78. To find the horse power required to move 
a. ear along a level track multiply together the dis- 
tance in feet traveled per minute and the force in 
pounds necessary to move the car (as found by Rule 
77), and divide the result by 33,000 times the ef- 
ficiency of the motors. 

Example. 

What horse power is needed to propel a loaded car 
weighing 9 tons along a level track at the rate of 
800 feet per minute, with motors of 70 per cent. 
efficiency ? 

Solution : Force to move car is 9 X 30 = 270 
pounds. The product of 800 X 270 = 216,000 foot 
pounds per minute. Dividing this by 33,000 gives 
6.54 H. P. required to propel the car. Dividing by 
the efficiency .70 gives 9.34 H. P. to be delivered to 
the motors. 

It will be noted in this solution that the quantity 
216,000 should, according to the rule, have been di- 
vided by the product of 33,000 times .70; it was, 



ELECTRIC RAILWAYS. 113 

however, divided by these two factors successively in 
order to show the difference between the power actu- 
ally moving the car and that supplied to the motors. 

In computing the power taken by a car ascending 
a grade the equivalent perpendicular rise of the car 
together with its weight in pounds have to be consid- 
ered in addition to the factors involved in the rule 
just preceding. 

Rale 79. To find the horse power required to pro- 
pel a car np a grade, take the product of the 
perpendicular distance in feet ascended by the car 
in one minute multiplied by its weight in pounds; 
to this add the product of the horizontal distance 
in feet traveled in one minute multiplied by the 
force in pounds required to propel the car; divide 
this sum by 33,000 times the efficiency of the mo- 
tors, 

Note. — The grade of a road or track is generally 
stated as being so many per cent. This means that 
for any given horizontal travel of a car its change of 
altitude when referred to a fixed horizontal plane is 
expressed as a certain percentage of the horizontal 
travel. For illustration; if a car while traveling 
horizontally 100 feet has a total vertical rise (or 
fall) of 7 feet, the incline on which it moves is 
termed a 7 per cent grade. 

Example. 
Find the electrical horse power taken by the motors 
of an 8 ton car to propel it up a 5 per cent grade at 



114 ARITHMETIC OF ELECTRICITY. 

a speed of 1,000 feet per minute, the motors having 
70 per cent efficiency. 

Solution: Perpendicular rise of car is 1,000 feet 
X .05 = 50 feet. Weight of car in pounds is 8 X 
2,000 = 16,000 pounds. Product of lift and weight 
is 50 X 16,000 = 800,000 foot pounds. Force re- 
quired to propel car is 30 X 8 = 240 pounds. Pro- 
duct of force and distance is 240 X 1,000 =240,000 
foot pounds. Sum of the two products is 800,000 -f- 
240,000 = 1,040,000 total foot pounds. 

Product of 33,000 by efficiency is 33,000 X .70 = 
23,100. Electrical H. P. is the quotient of 1,040,000 
-f- 23,100 = 45.021 H. P. Ans. 



CHAPTER XL 

ALTERNATING CURRENTS. 

By far the greater part of calculations in the do- 
main of alternating currents lie in the realm of 
trigonometry and the intricacies of the calculus. 
On this account it is hoped that the following pres- 
entation of some of the simpler formulae may prove 
welcome to the craft. 

A current flowing alternately in opposite directions 
may be considered as increasing from zero to a 
certain amount flowing in, say, the positive direc- 
tion, then diminishing to zero and increasing to an 
equal amount flowing in the negative direction and 
again decreasing to a zero value. This action is 
repeated indefinitely. The sequence of a positive 
and negative current as just described is called a 
cycle. 

The frequency of an alternating current is the 
number of cycles passed through in one second. An 
alternation is half a cycle. That is to say, an alter- 
nation may be taken as either the positive or the 
negative wave of the current. 



116 ARITHMETIC OF ELECTRICITY. 

The frequency may be expressed not only in cycles 
per second but in alternations per minute. 

Since one cycle equals two alternations we can 
interchange these expressions as follows: 

Rule 80. A. Having given the cycles per second, 
to find the alternations per minute multiply the 
cycles per second by 120, B. Having given the al- 
ternations per minute, to find the cycles per second 
divide the alternations per minute by 120, 

Examples. 

If a current has 60 cycles per second, how many 
alternations are there per minute? 

Solution: 60X120 = 7,200 alternations. 

A current has 15,000 alternations per minute; 
how many cycles per second are there ? 

Solution: 15,000-^-120 = 125 cycles per second. 

A bipolar dynamo having an armature with but 
a single coil wound upon it (like an ordinary mag- 
neto generator) gives one complete cycle of current 
for every revolution of the armature. That is to 
say, its frequency equals the number of revolutions 
per second. A four-pole generator will have a fre- 
quency equal to twice the revolutions per second, etc. 

Rnle 81. To find the frequency of any alternator, 
divide the revolutions per minute by 60 and multiply 
the quotient by the number of pairs of poles in the 
field. 

Example. 
Find the frequency of a 16-pole alternator run- 
ning at 937.5 revolutions per minute. 



ALTERNATING CURRENTS. 117 

Solution: 937.5 -f- 60 = 15.625 rev. per second. 
15.625 X 8 = frequency of 125 cycles per second. 

Electrical measuring instruments used on alternat- 
ing currents do not indicate the maximum volts or 
amperes of such circuits, but the effective values are 
what they show. These effective values are the same 
as those of a continuous current performing the 
same work. 

Rule 82. The maximum volts or amperes of an 
alternating current may be found by multiplying; 
the average volts or amperes by 1.11. Reciprocally, 
the average values can be found by taking .707 
times the maximum values. 

Note that these figures are strictly true only for 
an exactly sinusoidal current. 

Example. 
Find the maximum pressure ' of an alternating 
current of 55 volts. 

Solution: 55 X 141 = 61.05 volts. Ans. 

Self-Induction*. 
In an alternating current circuit the flow of a 
current under a given voltage is determined not 
only by the resistance of the conductor in ohms but 
also by the self-induction of the circuit. Suppose a 
current to start at zero and increase to 10 amperes in 
a coil of 1,000 turns of wire. This -magnetizing 
force, growing from zero to 10,000 ampere-turns, 
surrounds the coil with lines of force whose action 



118 ARITHMETIC OF ELECTRICITY. 

upon the current in the coil is such as to resist its 
increase. Conversely, when the current is decreas- 
ing from 10 amperes to zero, the lines of force 
change their direction and tend to prolong the flow 
of current. This opposing effect which acts on a 
varying or an alternating current is caller the counter 
E. M. F. or E. M. F. of self-induction and is meas- 
ured in volts. 

Rule 83. The E. M. F. of self-induction of a given 
coil is found by multiplying together 12.5664, the 
total number of turns in the coil, the number of 
turns per centimeter length of the coil, the sec- 
tional area of the core of the coil in square centi- 
meters, the permeability of the magnetic circuit 
and the current; divide the resulting product by 
1,000,000,000, multiplied by the time taken by the 
current to reach its maximum value* 

Example. 

Find the volts of counter E. M. F. in a coil of 300 
turns wound uniformly on a ring made of soft iron 
wire, the ring having a mean circumference of 60 
centimeters and an effective sectional area of 25 
square centimeters; its permeability to be taken as 
200, and a current of 5 amperes in the coil requires 
.02 second to reach its maximum value. 

Solution: Product of 12.5664 X 300 X 5 X 25 
X 200 X 5 is 471,240,000. Product of 10 9 X .02 is 
20,000,000. Dividing the former product by the 
latter gives 23.562 volts, Answer. 

The coefficient of self-induction or, as it is more 



ALTERNATING CURRENTS. 119 

frequently termed, the inductance of a coil, is meas- 
ured by the number of volts of counter E. M. F. 
when the current changes at the rate of one ampere 
per second. (Atkinson.) 

The unit of inductance is the henry. 

Rule 84. The inductance of a coll In found by 
multiplying together 12.5664, the total number of 
turns in the coil, the number of turns per centimeter 
length, the sectional area of the core of the coil in 
square centimeters, and the permeability of the 
magnetic circuit) divide the resulting product by 
1,000,000,000. 

Example. 
t 

Find the inductance of the coil specified in the 

preceding example. 

Solution: The factors are the same as before, 
omitting the current and time. Product of 12.5664 X 
300 X 5 X 25 X 200 is 94,248,000. Dividing this 
by 1,000,000,000 gives the inductance .094248 
henrys. 

The E. M. F. of self-induction may be computed 
when the inductance, the current and the time taken 
for the current to reach its maximum are known. 

Rule 85. To find the E. M. F. of self-induction 
divide the product of the inductance and current by 
the time of current rise* 

Example. 

Using again the data of the foregoing examples, 
find the counter E. M. F., the inductance being 



120 ARITHMETIC OF ELECTRICITY. 

.094248 henrys, the current 5 amperes, and the time 
.02 second. 

Solution: 5 X .094248 = .47124; dividing this 
by .02 gives 23.562 volts, as before. 

Rule SO. The resistance clue to self-induction 
equals 6.2832 times the product of the frequency 
and the inductance. 

Example. 

Find the inductive resistance of a circuit whose 
frequency is 60 cycles per second and the inductance 
is .05 henry. 

Solution: 6.2832 X 60 X .05 = 18.8496 ohms. 
Ans. 

The time constant of an inductive circuit is a 
measure of the growth or increase of the current. 
It is the time required by the current to rise from 
zero to its average value. The average value of an 
alternating current is .634 times its maximum value. 
It must not be confused with its effective value, 
which is .707 times the maximum. 

The average value may be obtained by multiplying 
the effective value, as shown by instruments, by .897. 

Rule 87. To find the time constant of a coil or 
circuit, divid© its Inductance by its resistance* 

Example. 
What is the time constant of a coil whose induct- 
ance is 3.62 henrys and resistance is 20 ohms. 
Solution: 3.62 -=-20 = .181 second. Ans. 



CHAPTER XII. 

CONDENSERS. 

A condenser, though it will allow no current to 
pass through it, yet it will accumulate or store up 
a quantity of electricity depending on various factors 
which the following rules will show: 

Role 88. The quantity stored equal* the product 
of the E. M. P. applied and the capacity of the 
condenser. 

Q, = EC, 

Rule 89. The capacity of a condenser equals the 
quantity stored divided by the applied E. M. F. 

a 

E 

Rule 90. The E. M. P. applied to a condenser equals 
the quantity stored divided hy its capacity. 

Q, 

E = — . 
C 

The quantity stored in a condenser is measured 
in coulombs (i. e., ampere-seconds) ; the E. M. F. 
in volts, and the capacity in farads. Condensers in 
practical use have, however, so small a capacity that 



122 ARITHMETIC OF ELECTRICITY. 

it is usually stated in microfarads and the quantity 
in microcoulombs. 

Examples. 

A battery of 30 volts E. M. F. is connected to a 
condenser whose capacity is one half microfarad. 
What quantity of electricity will be stored? 

Solution: 30 volts X .0000005 farads = .000015 
coulombs. This solution could also be given direct- 
ly in micro-quantities, thus: 30 volts X V2 micro- 
farad = 15 microcoulombs. 

A condenser is charged with 7.5 microcoulombs 
under an E. M. P. of 15 volts. What is its capacity? 

Solution: 7.5 microcoulombs -f- 15 volts = .5 
microfarad. Ans. 

What E. M. F. is required to charge a condenser 
whose capacity is .1 microfarad with 21 microcoul- 
ombs of electricity? 

Solution: 21 microcoulombs -r- .1 microfarad = 
210 volts. Ans. 

By connecting condensers in parallel the resulting 
capacity is the sum of their individual capacities. 
When they are connected in series the resulting capa- 
city equals 1 divided by the sum of the reciprocals 
of their individual capacities. It will be noticed that 
these laws of condenser connections are the inverse 
of those for the parallel and series connection of re- 
sistances. 



CONDENSERS. 123 

When applying a direct current to a condenser, 
as in the above examples, it flows until the increasing 
charge opposes an E. M. F. equal to that of the 
charging current. 

With an alternating current a charge would be 
surging in and out of the condenser, so that a real 
current will be flowing on the charging wires in 
spite of the fact that the actual resistance of a con- 
denser, in ohms, is practically infinite. 

Rule 91. The alternating current in a circuit hav- 
ing: capacity equals the product of 6.2832, the fre- 
quency, the capacity, and the applied voltage* 

Example. 

Find the current produced by an E. M. P. of 50 
volts and a frequency of 60 cycles per second in a 
circuit whose capacity is 125 microfarads. 

Solution: The capacity 125 microfarads equals 
.000125 farads. 

6.2832 X 60 X .000125 X 50 = 2.3562 amperes. 
Ans. 

Rule 92. The alternating E. M. P. required to he 
impressed upon a circuit of a given capacity in 
order to produce a certain current is equal to the 
current divided by 6.2832 times the product of the 
capacity and the frequency. 

Example. 

Find the E. M. F. necessary to produce an alter- 
nating current of 50 amperes at 50 cycles per sec- 
ond in a circuit of 80 microfarads capacity. 

Solution: 6.2832 X .000080 X 50 = .0251328 



124 ARITHMETIC OF ELECTRICITY. 

Dividing the current 50 amp. by .0251328 gives 2,000 
volts, nearly. 

Since, in a condenser circuit, a real current flows 
under a given E. M. F., the circuit may be treated as 
though it was of a resistance such as would allow 
the given current to flow. 

Rule 93. The reHiHtance due to capacity equals 1 
divided by the product of G.28^2, the frequency and 
the capacity. 

Example. 

Find the capacity resistance of a circuit having a 
frequency of 60 cycles per second and a capacity of 
50 microfarads. 

Solution: 6.2832 X 60 X .000050 = .01885. 
1 -f- .01885 = 53 ohms, very nearly. 

By comparing this Eule 93 for capacity resistance 
with Eule 8G on page 120, which is for inductive re- 
sistance, it will be seen that they are mutually recip- 
rocal and hence the effect of capacity is directly 
opposite to that of self-induction and vice versa. 

It follows from this that it is possible, by the 
proper proportioning of the inductance and the 
capacity, to have their effects neutralized, and when 
this adjustment is effected the current will be con- 
trolled by the volts and ohmic resistance the same 
as if it were a direct current circuit. 

The impedance is the apparent resistance of an 
alternating current circuit. 



CONDENSERS. 125 

Rule 94. To find the impedance of a circuit whose 
olimic resistance can be neglected and which has 
an inductance and a capacity in series, calculate 
botli the inductive resistance and the capacity re- 
sistance; their difference will be the impedance. 

Example. 

Find the current produced by an alternating E. 
M. P. of 40 volts on a circuit of slight ohmic resist- 
ance whose capacity is 100 microfarads, the frequency 
being 60 cycles per second, and having in series an 
inductance of .02 henry. 

Solution: Inductive resistance is 6.2832 X 60 
X .02 = 7.42 ; capacity resistance is 1 -~ 6.2832 X 
60 X .000100 = 1 ~- .0377 = 26.52. 

Impedance = 26.52 — 7.12 = 19.1 ohms. 

Current, by Ohm's Law, = 40 -f- 19.1 = 2.08 am- 
peres. Ans. 



CHAPTEK XIII. 

DEMONSTRATION OF RULES. 

In the following chapter we give the demonstra- 
tion of some of the rules. As this is not within the 
more practical portion of the work, algebra is used 
in some of the calculations. It is believed that rules 
not included in this chapter, if not based on experi- 
ment, are such as to require no demonstration here. 

Rule 1 to 6, pages 13 and 14. Ohm's law was 
determined experimentally, and all the six forms 
given are derived by algebraic transposition from 
the first form which is the one most generally ex- 
pressed. 

Rule 8, page 19. This is simply the expression of 
Ohm's law as given in Eule 1, because in the case of 
divided circuits branching from and uniting again 
at common points, it is obvious that the difference 
of potential is the same for all. Hence the ratio as 
stated must hold. 

Rule 9 3 page 20. This rule is deduced from Kule 
8. It first expresses by fractions the relations of 
the current. Next these fractions are reduced to a 
common denominator, so as to stand to each other in 



128 ARITHMETIC OF ELECTRICITY. 

the ratio of their numerators. By applying the new 
common denominator made up of the sum of 
the numerators the ratio of the numerators 
is unchanged, and the ratio of the new fractions is 
the same as that of their numerators, while by this 
operation the sum of the new fractions is made 
equal to unity. Thus by multiplying the total cur- 
rent by the respective fractions it is divided in the 
ratio of their numerators, which are in the inverse 
ratio of the resistances of the branches of the circuit 
and as the sum of the fractions is unity, the sum of 
the fractions of the current thus deduced is equal to 
the original current. 

Rule 10, page 22. Resistance is the reciprocal of 
conductance. By expressing the sum of the recip- 
rocals of the resistances of parallel circuits we ex- 
press the conductance of all together. The recipro- 
cal of this conductance gives the united resistance. 

Rule n, page 22. This is a form of Eule 10. 
Call the two resistances x and y. The sum of their 
reciprocals is \-\-\ which is the conductance of the 
two parallel circuits or parts of circuits. Reducing 
them to a common denominator we have: X + iL 

xy xy 

which equals ^r^, whose reciprocal is ~r-* 

Rule 17, page 31. Taking the diameter of a wire 
as d, its cross sectional area is ^-. The resistance 
is inversely proportional to this or varies directly 
with T^^-" 1 ^-. ^- s ^ ie distance of a conductor 



DEMONS TBA TION OF B ULES. 120 

varies also with its length and specific resistance we 
have as the expression for resistance: 

Sp. Res, x 1.2737 Xl 

Rule 18, page 32. Assume two wires w r hose 
lengths are I and l v their cross sectional areas a and 
a v their specific resistances s and * l3 and their resis- 
tances r and r v From preceding rules we have for 
each wire: r ■=* i (1) and r% = s x ~ (2). 

Dividing (1) by (2) we have: 

k = k. X h X Z<?} (Day) 

If we take the reciprocal of either member of this 
equation and multiply the other member thereby it 
will reduce it to unity, or: 

For convenience this is put into a shape adapted 
for cancellation. 

Rule 20, page 38. This is merely the expression 
of Ohm's Law, Rule 3. 

Rule 22, page 40. Call the drop e, the combined 
resistance of the lamps R, and the resistance of the 
leads x. Then as the whole resistance is expressed 
as 100 (because the work is by percentage) the differ- 
ence of potential for the lamps is 100 — e. By 
Ohm's law we have the proportion: 100-6 : e :: R : 
x or 

eR 
■ = 10O-« 

Rule 25, page 44. From Rule 22 we have: 



eR 
: 100— e 



(i) 



130 ARITHMETIC OF ELECTRICITY. 

Call the resistance of a single lamp r, then we 
have by Eule 12: 

R= ,T (2) 

Substituting this value of R in equation (1) we 
have: 

— e r 

X "~ n X (100— e) (3) 

Prom Kule 24 we have, calling the cross-section a: 

l X 10.79 

*=— T- (4) 

Substituting for x its value from equation (3) we 
have: 

» ^XlO.79 X n X (100-e) 

— - r (5) 

But as I expresses the length of a pair of leads, 
not the total length of lead but only one-half the 
total, the area should be twice as great. This is 
effected by using the constant 10.79 X 2 == 21.58 in 
the equation giving: 

l X 21.58 X n X (100— e) 

a = 

er 

Rule 28, page 48. Assuming the converter to work 
with 100$ efficiency (which is never the case), the 
watts in the primary and secondary must be equal 
to each other or: 

C 2 R = (V B lf and R = %? R x , 

or the resistances of primary and secondary are in 
the ratio of the squares of the currents. The direct 
ratio is expressed by the ratio of conversion, when 
squared it gives the ratio of the squares as required. 



DEMONSTRATION OF RULES. 131 

Rule 37, page 59. Let d = diameter of the wire in 
centimeters. The resistance of one centimeter of 
such a wire in ohms = Sp. Resist. X 10" 6 X ~^. The 
specific resistance is here assumed to be taken in 
microhms. The quantity of heat in joules devel- 
oped in such a wire in one second is equal to the 
square of the current in C. G. S. units, multiplied 
by the resistance in C. G-. S. units and divided by 
4.16 X 10 7 , the latter division effecting the reduc- 
tion to joules. 1 ohm = 10 9 C. G. S. units of re- 
sistance. Multiplying the expression for ohmic 
resistance by 10 9 we have: Sp. Resist. X 10 8 X ^~, 
1 ampere = 10 1 0. G. S. unit. If we express the 
current in amperes we must multiply it by 10* x , in 
other words take one-tenth of it. Our expression 
then becomes for heat developed in one second 



tt> 



y Sp. Resist. X 10 3 X 4 
v d 2 X 4.16 X 10* 



The area of one centimeter of the wire is *■ d 
square centimeters. The heat developed per square 
centimeter is found by dividing the above expression 
by nd giving: 

fJL\ 2 x Sp. Resist. X 10 8 X 4 
\\0) Tr^d 3 X 4.16 X 10 7 

The heat developed is opposed by the heat lost 
which we take as equal to t^ts per square centimeter 
per degree Cent, of excess above surrounding me- 
dium. Therefore taking t° as the given tempera* 



132 ARITHMETIC OF ELECTRICITY. 

ture cent, we may equate the loss with the gain 
thus: 

t° /_£\ a x S P Resist. X 10 3 X 4 

4000 " V10/ «r« d 3 X 4.16 X 10 7 

_ C 2 X Sp. Resist. X 10 8 X 4 X 40000 _ 
*» X 4.16 X 10 7 X t° " 

c* x Sp. Resist, x .00039 

t° 

iiuie52 9 page 69. Call the external resistance r; 
number of cells n; resistance of one cell R; E, M. 
F. of one cell E; E. M. F. of outer circuit e. 

Then from Ohm's law we have: 

n_ nE 

~ nR + r (1) 



which reduces to: 
but O r = e. 



Or 

e—cr (2) 



•'• n " W=TCR (3) 

Rnic 54, page 72. This rule is deduced from the 
following considerations. The current being con- 
stant the work expended in the battery and external 
circuit respectively will be in proportion to their 
differences of potential or E. M. F's. But these 
are proportional to the resistances. Therefore the 
resistance of the external circuit r should be to the 
resistance of the battery R as efficiency: 1 — effi- 
ciency or r : R :: efficiency : 1 — efficiency or 

R = (1 ~ e eS C clen C cy )Xr ' The rest ° f the rule is Educed 

from Ohm's law. 



DEMONSTRATION OF RULES. 133 

Rule 57, page 74. This rule gives the nearest ap- 
proximation attainable without irregular arrange- 
ment of cells. By placing some cells in single series 
and others two or more in parallel, an almost exact 
arrangement for any desired efficiency can be ob- 
tained. Such arrangement are so unusual that it is 
not worth while to deduce any special rule for 
them. Thus taking the example given on page 
74 the impossible arrangement of 1.4 cells in 
parallel and 63 in series would give the desired 
current and efficiency. The same result can 
be obtained by taking 72 cells in 36 pairs with a re- 
sistance of 36 x A = 3 ohms, and adding to them 
27 cells in series with a resistance of 27x| = 4£ 
ohms, a total of 7i ohms. The E. M. P. is equal to 
(36 + 27) X 2 = 126 volts. The total cells are 
72 + 27 = 99. 

Rule 58, page 76. One coulomb of electricity lib- 
erates from an electrolyte .000010384 gram of 
hydrogen. This has been determined experimen- 
tally. Let H be the heat liberated by the chemical 
combining weight of any body combining with 
another. H is taken in kilogram calories. Hence it 
follows that for a quantity of the substance equal to 
.000010384 gram X chemical combining weight, the 
heat liberated will be equal to H X .000010384, 
which corresponds to a number of kilogram meters 
of work expressed by .000010384 X H X 424. The 
work done by a current in kilogram-meters = 



134 ARITHMETIC OF ELECTRICITY. 

volts X coulombs _ # i r. volts mi • 

£gj or for one coulomb = ~. This ex- 
presses the work done by one coulomb. Let the 
volts = E, and equate these two expressions: 
-j = .000010384 X H X 424, 
which reduces to 

E = H X .043. 

Rule 6i, page 78. For the work (in kilogram- 
meters) done by a current (volt-coulombs) we have 
the general expression: 

_ volts x coulombs AM EQ 

w= m — or m (i) 

Making W =* 1 (i. e. one kilogram-meter) and 
transforming, we have, as the coulombs correspond- 
ing to 1 kilogram-meter: 

n 9.81 

Q= -E" (2) 

One coulomb of electricity liberates a weight (in 
grams) of an element equal to the product of the 
following: .000010384 X equivalent of element in 
question X number of equivalents +■ valency of the 
element. Therefore, the coulombs corresponding 
to one kilogram-meter, liberates this weight multi- 
plied by ~ or, indicating weight by G, 

a - -000010384 x equiv. x number equiv. x 9L81 

valency F. (3) 

but .000010384 X 9.81 = .000101867. 

__ equiv. X n X .000101867 

"" EX valency (4) 

Rale 73, page 98-99. The voltage of an armature of 



DEMONSTRATION OF RULES. 135 

a definite number of turns of wire and a fixed speed, 
varies with the lines included within its longitu- 
dinal area, as such lines are cut in every revolution. 
These lines vary with its area, and the latter varies 
with the square of its linear dimensions. 

To maintain a constant voltage if the size is 
changed, the number of turns must be varied in- 
versely as the square of the linear dimensions. 
This ensures the cutting of the same number of 
lines of force per revolution. 

If, therefore, its size is reduced from x to \ the 
turns of wire must be changed from x to &. The 
relative diameters of the two sizes of wire is found 
by dividing a similar linear dimension by the rela- 
tive size of the wire. But 1 -*• x 2 = ~ = diameter of 
the wire for maintenance of a constant voltage with 
change of size. 

The capacity of a wire varies with the square of 
its diameter and (±y -. \p 

Therefore the amperage, if a constant voltage is 
maintained, will vary inversely as the sixth power of 
the linear dimensions of an armature. 



CHAPTER XIV. 

NOTATION IN POWEES OF TEN. 

This adjunct to calculations has become almost 
indispensable in working with units of the C. G. S. 
system. It consists in using some power of 10 as a 
multiplier which may be called the factor. The 
number multiplied may be called the characteristic. 
The following are the general principles. 

The power of 10 is shown by an exponent which 
indicates the number of ciphers in the multiplier. 
Thus 10 2 indicates 100; 10 3 indicates 1000 and so 
on. 

The exponent, if positive, denotes an integral 
number, as shown in the preceding paragraph. The 
exponent, if negative, denotes the reciprocal of the 
indicated power of 10. Thus 10~ 2 indicates jh; 10 3 
indicates toW and so on. 

The compound numbers based on these are re- 
duced by multiplication or division to simple expres- 
sions. Thus: 3.14 X 10 7 = 3.14 X 10,000,000 = 
31,400,000. S.UXlO-^^or ^^ . Ee- 
gard must be paid to the decimal point as is done 
here. 



NOTATION IN POWERS OF TEN. 137 

To add two or more expressions in this notation 
if the exponents of the factors are alike in all re- 
spects, add the characteristics and preserve the same 
factor. Thus: 

(51 X 10 6 ) + (54 X 10* )= 105 X 10 6 . 
(9.1 X 10" g ) + (8.7 X 10" 9 ) = 17.8 X 10" 9 . 

To subtract one such expression from another, 
subtract the characteristics and preserve the same 
factor. Thus: 

(54 X 10 6 ) - (51 X 10 8 ) = 3 X 10 6 . 

If the factors have different exponents of the 
same sign the factor or factors of larger exponent 
must be reduced to the smaller exponent, by factor- 
ing. The characteristic of the expression thus 
treated is multiplied by the odd factor. This gives 
a new expression whose characteristic is added 
to the other, and the factor of smaller exponent is 
preserved for both. 

Thus: 

(5 X 10 7 ) + (5 X 10 9 ) = (5 X 10 7 ) + (5 X 100 X 
10 7 ) = 505 X 10 7 . 

The same applies to subtraction. Thus: 

(5 X 10^) - (5 X 10 7 ) = (5 X 100 X 107) - (5 X 
10 7 ) = 495 X 10 7 . 

If the factors differ in sign, it is generally best to 
leave the addition or subtraction to be simply ex- 



138 ARITHMETIC OF ELECTRICITY. 

pressed. However by following the above rule it 
can be done. Thus: 

Add 5 X 10- 2 and 5 X 10 3 . 

5 X 10 8 = 5 X 10 5 X lO" 2 : (5 X 10 5 X 10" 2 ) + (5 X 
10"*) = 500005 X 10* 2 . This may be reduced to a 
fraction 6 -^= 5000.05. 

To multiply add the exponents of the factors, for 
the new factor, and multiply the characteristics for 
a new characteristic. The exponents must be added 
algebraically: that is, if of different signs the numer- 
ically smaller one is subtracted from the other one, 
its sign is given the new exponent. 

Thus: 

(25 X 10 6 ) X (9 x 10 8 ) = 225 X 10 u . 
(29 X 10 8 ) X (11 X 10 7 ) = 319 X 10 1 . 
(9 X 10 8 ) X (98 X lO" 2 ) = 882 X 10 6 . 

To divide, subtract (algebraically) the exponent 
of the divisor from that of the dividend for the ex- 
ponent of the new factor, and divide the character- 
istics one by the other for the new characteristic. 
Algebraic subtraction is effected by changing the 
sign of the subtrahend, subtracting the numer- 
ically smaller number from the larger, and giving 
the result the sign of the larger number. (Thus to 
subtract 7 from 5 proceed thus: 5 — 7 = —2.) 

Thus: 

(25 X 10 6 ) +(5X 10 8 ) - 5 X lO" 8 
(28 X lO 8 ) + (5 X 10 8 ) = 5.6 X 10" 11 . 



TABLES. 



*g 






I 






B 



5 



s 






B 















o 



H 

GD 
O 

w 

Q 

w 



8 
§ 



p 



140 
II.-EQUIVALENTS OF UNITS OF AREA. 





Square 
Millimeter 


Square 
Centimet'r 


Circular 
Mil. 


Square 
Mil. 


Square 
Inch. 


Square 
Foot. 


Square Millimeter 


1 


0.01 


1973.6 


1550.1 


.00155 


.0000108 


Square Centimeter 


100 


1 


197,361 


155,007 


.155007 


.C01076 


Circular Mil. 


.000507 


.0000051 


1 


.7S540 


8X10-' 




Square Mil. 


.000645 


.0000065 


1.2733 


1 


.000001 




Square Inch 


645.132 


6.451 


1,273,238 


1,000,000 


1 


.006944 


Square Foot 


*%898.9 


928.989 






144 


1 





m.~ 


EQUIVALENTS OF UNITS OF VOLUME. 






Cubic 
Inch 


Fluid 
Ounce 


Gallon 


Cubic 
Foot 


Cubic 
Yard 


Cu. Cen- 
timeter 


Liter 


Cubic 
Meter 


Cubic Inch 

-* 


1 


.554112 


.004329 


.000578 




16.8862 


.016386 




Fluid Oz. 


1.80469 


1 


.007812 


.001044 




29.5720 


.029572 




Gallon 


231 


128 


1 


.183681 


.00495 


8785.21 


8.78521 


.003785 


Cubic Ft. 


1728 


957.506 


7.48052 


1 


.037087 


28315.8 


28.8153 


.028315 


Cubic Yd. 


46,656 


25,852.6 


201.974 


27 


1 


764,505 


764.505 


.764505 


Cu. Centi. 


.061027 


.033816 


.000264 


.000035 




1 


.001 


.000001 


Liter 


61.027 


33.8160 


.2641S9 


.035317 




1000 


1 


.001 


Cu. Meter 


61027 


33S16 


204.189 


35.3169 


1.80S0 




1000 


1 



141 
IV.— EQUIVALENTS OF UNITS OF WEIGHT. 





Grain. 


Troy 
Ounce. 


Pound 

Avs. 


Ton. 


Milli- 
gram. 


Gram. 


Kilo- 
gram. 


Metric 
Ton. 


Grain 


1 


.020833 


.000143 




64.799 


.064799 


.000065 




TroyOunce 


480 


1 


.068641 




31,103.5 


31.1035 


.031104 




PoundAvs. 


7,000 


14.5833 


1 


.000447 




453.593 


.453593 


.000454 


Ton 




32,666.6 


2240 


1 






.001016 


1.01605 


Milligram 


.015432 


.000032 


.000002 




1 


.001 


.000001 




Gram 


15.4323 


.032151 


.002205 




1000 


1 


.001 




Kilogram 


15,432.3 


32.150T 


2.20462 


.0009&4 


1,000,000 


1000 


1 


.001 


Metric Ton 




32,150.7 


2204.62 


.98421 




1,000,000 


1000 


1 



142 



V.-EQTJIVALENT8 OF UNITS 





Erg. 


Meg- 
erg. 


Gram-de- 
gree C. 


Kilogram- 
degree C. 


Pound- 
degree C. 


Pound- 
degree F. 


Erg. 


1 


.000001 










Meg. -erg. 


1,000,000 


1 


.024068 


.000024 


.000053 


.000095 


Gram-degree C. 




41.5487 


1 


.001 


.002205 


.003968 


Kilogram -degree C. 




41,548.7 


1000 


1 


2.2046 


8.9688 


Pound-degree C. 




18,846.5 


453.59 


.45359 


1 


1.8 


Pound -degree F. 




10,470.1 


251.995 


.251995 


.555556 


*\ 


Watt-Second. 


10 7 


10 


.24068 


.000241 


.000531 


.000955 


Gram-centimeter. 


981 


.000981 


.0000235 








Kilogram-meter. 


98.1X10« 


98.1 


2.86108 


.002861 


.005205 


.009370 


Foot-Pound. 




13.5626 


.826425 


.000826 


.000720 


.001295 


Horse-Power-Sec. 

English. 




7459.48 


179.486 


.179486 


.8957 


.71243 


Horse-Power-Sec. 
Metric. 




7357.5 


177.075 


177.075 


.890375 


.70275 



143 



OF ENERGY AND WORK. 



Watt- 
Second. 


Gram- 
Centim'tr. 


Kilogram- 
meter. 


Foot- 
pound. 


Horse- 
power- 
second 
English. 


Horse- 
power- 
second 
Metric. 




10- r 


.001019 










Erg. 


.1 


1019.87 


.010194 


.073734 


.000134 


.000136 


Meg-erg. 


4.15487 


42,853.5 


.423535 


3.06355 


.00557 


.005647 


Gram-degree C. 


4154.87 




423.535 


3063.55 


5.57 


5.64703 


Kilogram-degree C. 


18S4.65 




192.114 


1389.6 


2.52653 


2.56149 


Pound-degree C. 


1047.03 




106.730 


772 


1.40364 


1.42305 


Pound-degree F. 


1 


10,193.7 


.101937 


.737337 


.0018406 


.0018592 


Watt-Second. 


.000098 


1 


.00001 


.000072 






Gram-Centimeter. 


9.81 


100,000 


1 


7.28328 


.013152 


.013334 


Kilogram-meter. 


1.35626 


13,825.8 


.138253 


1 


.0018182 


.001843 


Foot-Pound. 


745.943 




76.0392 


550 


1 


1.01888 


Horse-Power-Sec. 

English. 


735.75 




75 


542.496 


.986356 


1 


Horse-Po wer-S ec . 
Metric 



144 



VI.— TABLE OF SPECIFIC RESISTANCES IN MICROHMS AND 
OF COEFFICIENTS OF SPECIFIC RESISTANCES OF METALS. 





Specific 


Coeffi- 




Specific 


Coefin- 




Resist- 


cients of 




Resist- 


cients of 




ance. Mi- 
crohms. 


Sp. Res. 




ance. Mi- 
crohms. 


Sp. Res. 


Annealed Silver 


1.521 


.9412 


Annealed Nickel 


12.60 


7.7970 


Hard Silver 


1.652 


1.0223 


Compres'd Tin 


13.36 


8.2673 


Annealed Copper. . . . 


1.616 


1.0000 


" Lead . . 


19.85 


12.2834 


Hard Copper 

Annealed Gold... .. 


1.652 


1.0223 


" Antimony 


35.90 


22.2153 


2.0S1 


1.2877 


" Bismuth . 


132.70 


82.1170 


Hard Gold 


2.118 


1.3107 


Liquid Mercury . . . 


99.74 


61.7203 


Annealed Aluminum.. 


2.945 


1.8224 


2 Silver, 1 Platinum. 


24.66 


15.2599 


Compressed Zinc 


5.6S9 


3.5204 


German Silver 


21.17 


13.1002 


Annealed Platinum . . 


9.158 


5.6671 


2 Gold, 1 Silver .... 


10.99 


6.8008 


" Iron 


9.825 


6.0798 









SPECIFIC RESISTANCE OF SOLUTIONS AND LIQUIDS. 

MATTHIE8SEN AND OTHEB8. 



Names of Solutions. 


Temper- 
ature 
Centi- 
grade. 


Temper- 
ature 

Fahren- 
heit. 


Specific 

Resistance. 

Ohms. 


Copper Sulphate, concentrated 


9° 
it 

it 

13° 
(t 

it 

it 

1 4 q 
ii 

14.3° 
14.5° 
12.3° 

12.8« 
14° 
24° 


48.2° 
ii 

ti 
55.4° 

ti 
i. 

57.2° 
it 

ii 
57.8° 
58.1° 
54.5° 

55.0° 
57.2° 
75.2° 


29.82 


u with an equal volume of water 

" with three volumes of water 

Common Salt, concentrated 


46.54 

77.68 

5.93 


" with an equal volume of water. . 

M with two volumes of water 

11 with three volumes of water. . 
Zinc Sulphate, concentrated 


6.00 

9.24 

11.89 

28.00 


" with an equal volume of water . . 

M with two volumes of water 

Sulphuric Acid, concentrated 


22.75 

29.75 

5.32 


41 50.5#, Specific Gravity 1.393 . . . 

" 29.6*, Specific Gravity 1.215. . . 

" 12* Specific Gravity 1.0S0... 

Nitric Acid, Specific Gravity 1.36 (Blavier) .... 

it «c « it 


1.086 
.83 
1.368 
1.45 
1.22 


Distilled Water, (Temp'ture unknown) (Pouillet) 


932. 



14b 



VII.— RELATIVE RESISTANCE AND CONDUCTANCE OF PURE 
COPPER AT DIFFERENT TEMPERATURES. 

MATTHIES8EN. 



? © 


© 
J © 

si 


Relative 
Resistance. 


Relative 
Conductance 


Si 

cj OS 
*+ t" 

© fcp 

cu:p 

a a 

8 « 

16° 


© 

-C © 

03^=1 

h* a 

a| 


Relative 
Resistance. 


Relative 
Conductance 


0° 


82* 


1. 


1 


60.8Q 


1.06168 


.9419 


1 


33.8 


1.00381 


.99620 


17 


62.6 


1.06563 


.93841 


2 


35.6 


1.00756 


.9925 


18 


64.4 


1.06959 


.93494 


3 


87.4 


1.01135 


.98878 


19 


66.2 


1.07356 


.93148 


4 


89.2 


1.01515 


.9S508 


20 


68. 


1.07754 


.92804 


5 


41 


1.01896 


.98139 


21 


69.8 


1.08152 


.92462 


6 


42.8 


1.0228 


.97771 


22 


71.6 


1.08553 


.92120 


7 


44.6 


1.02663 


.97406 


23 


73.4 


1.08954 


.91782 


8 


46.4 


1.03048 


.97042 


24 


75.2 


1.09356 


.91445 


9 


48.2 


1.03435 


.96679 


25 


77. 


1.09759 


.9111 


10 


50 


1.03822 


.96319 


26 


78.8 


1.10162 


.90776 


11 


51.8 


1.04210 


.95960 


27 


80.6 


1.10567 


.90443 


12 


53.6 


1.04599 


.95603 


28 


82.4 


1.10972 


.90113 


13 


55.4 


1.0499 


.95247 


29 


84.2 


1.11382 


.89784 


14 


57.2 


1.05381 


.94898 


30 


86. 


1.11785 


.89457 


15 


59 


1.05774 


.94541 











146 



YIII.-AMERICAN WIRE GAUGE TABLE. 

Properties of Copper Wire : Speciflo Gravity, 8.8T8 ; Specific Conductivity, 1.765 at 75° F. 



1 

a 


Size. 


Weight and Length. 


Resistance. 


o d * 

fill 



ft 

1 

O 


Diam- 
eter in 
Mils. 


Square of 
Diameter 
or circular 

Mils. 


Grains 

per 

Foot. 


Po'nds 
per 
1000 
Feet. 


Feet 

per 

Pound. 


Ohms 

per 1000 

Feet. 


Feet 

per 

Ohm 


Ohms per 
Pound. 


* 2 i? a 


0000 


460.000 


211600.0 


4477.2 


639.60 


1.564 


.051 


19929.7 


.0000785 


430 


000 


409.640 


167804.9 


3550.5 


507.22 


1.971 


.063 


15804.9 


.000125 


262 


00 


864. S00 


133079.0 


2815.8 


402.25 


2.486 


.080 


12534.2 


.000198 


208 





824.950 


105592.5 


2236.2 


319.17 


8.188 


.101 


9945.3 


.000815 


165 


1 


289.300 


83694.49 


1770.9 


252.98 


8.952 


.127 


7882.8 


.000501 


130 


2 


257.680 


66373.22 


1404.4 
1118.6 


200.68 


4.994 


.160 


6251.4 


.000799 


103 


8 


229.420 


52633.53 


159.09 


6.285 


.202 


4957.8 


.001268 


81 


4 


204.810 


41742.57 


883.2 


126.17 


7.925 


.254 


8931.6 


.002016 


65 


6 


181.940 


83102.16 


700.4 


100.05 


9.995 


.821 


8117.8 


.003206 


52 


6 


162.020 


26250.48 


555.4 


79.34 


12.604 


.404 


2472.4 


.005098 


41 


7 


144.280 


20816.72 


440.4 


62.92 


15.893 


.509 


1960.6 


.008106 


82 


8 


128.490 


16509 . 68 


849.3 


49.90 


20.040 


.648 


1555.0 


.01289 


26 


9 


114.480 


13094.22 


277.1 


89.58 


25.265 


.811 


1238.3 


.02048 


20 


10 


101.890 


10381.57 


219.7 


31.38 


81.867 


1.028 


977.8 


.03259 


16 


11 


90.742 


8234.11 


174.2 


24.89 


40.176 


1.289 


775.5 


.05181 


13 


12 


80. 80S 


6529.93 


188.2 


19.74 


50.659 


1.62? 


615.02 


.08287 


10.2 


13 


71.961 


5178.39 


109.6 


15.65 


63.898 


2.048 


488.25 


.13087 


8.1 


14 


64.084 


4106.75 


86.87 


12.41 


80.580 


2.585 


886.80 


.20880 


6 4 


15 


57.068 


8256.76 


68.88 


9.84 


101.626 


8.177 


806.74 


.83183 


5.1 


16 


50.820 


2582.67 


54.67 


7.81 


128.041 


4.582 


248.25 


.52638 


4.0 


17 


45.257 


2048.19 


43.83 


6.19 


161.551 


5.183 


192.91 


.83744 


8.2 


18 


40.303 


1624.88 


84.87 


4.91 


203.666 


6.536 


152.99 


1.8812 


2.5 


19 


85.390 


1252.45 


26.50 


8.786 


264.186 


8.477 


117.96 


2.2392 


1.96 


20 


81.961 


1021.51 


21.60 


8.086 


824.045 


10.894 


96.21 


3.3488 


1.60 


21 


28.462 


810.09 


17.14 


2.448 


408.497 


18.106 


76.80 


5.8539 


1.28 


22 


25.847 


642.47 


18.59 


1.942 


514.938 


16.525 


60.51 


8.5099 


1.08 


23 


22.571 


509.45 


10.77 


1.539 


649.778 


20.842 


47.98 


13.884 


.80 


24 


20.100 


404.01 


8.55 


1.221 


819.001 


26.284 


88.05 


21.524 


.63 


25 


17.900 


820.41 


6.77 


.967 


1084.126 


88.185 


80.18 


84.298 


.50 


26 


15.940 


2M.08 


5.88 


.768 


1802.083 


41.789 


28.93 


54.410 


.40 


27 


14.195 


201.49 


4.26 


.608 


1644.787 


52.687 


18.98 


86.657 


.81 


28 


12.641 


159.79 


8.89 


.484 


2066.116 


66.445 


15.05 


187.283 


.25 


29 


11.257 


126.72 


2.69 


.884 


2604.167 


88.752 


11.94 


218.104 


.24) 


30 


10.025 


100.50 


2.11 


.802 


8811.258 


105.641 


9.466 


849.805 


.16 


81 


8.928 


79.71 


1.67 


.280 


4184.100 


133.191 


7.508 


557.286 


.18 


82 


T.950 


63.20 


1.38 


.190 


5268.158 


168.011 


5.952 


884.267 


.098 


88 


T.080 


50.13 


1.06 


.151 


6622.517 


211.820 


4.721 


1402.78 


.078 


84 


6.804 


89.74 


.847 


.121 


8264.468 


267.165 


8.748 


2207.98 


.062 


85 


5.614 


81.52 


.658 


.094 


10638.80 


836.81 


2.969 


8583.12 


.049 


86 


5.000 


25.00 


.525 


.075 


18333.83 


424.65 


2.855 


5661.71 


.039 


87 


4.453 


19.88 


.420 


.060 


16666.66 


535.33 


1.868 


8922.20 


.031 


88 


8.965 


15.72 


.815 


.045 


22222.22 


675.22 


1.481 


15000.5 


.025 


89 


3.531 


12.47 


.266 


.038 


26315.79 


851.789 


1.174 


22415.5 


.020 


40 


3.144 


9.88 


.210 


.030 


33383 38 


1074.11 


.931 


85803.8 


.015 



147 

IX.— CHEMICAL AND THERMO-CHEMICAL EQUIVALENTS. 

Formation of Oxides. 



Name of Compound. 


Formula. 


Valency. 


Chemical 
Equiv- 
alents. 


Combin- 
ing 
Weights. 


Thermo- 
Chemical 
Equiv- 
alents. 


Water 


H2 

FeO 

Fe2 03 

ZnO 

CuO 

HgO 


II 
II 
III 
11 
II 
II 


18 

72 
160 

81 

79.4 
216 


9 

86 
58.8 
40.5 
89.7 
108 


84.5 


Iron Protoxide 


84.5 


Iron Sesquioxide 


81.9x3 


Zinc Oxide 


48.2 


Copper Oxide 

Mercury Oxide 


19.2 
15.5 







Formation of Salts. 



Name of 
Base. 


Va- 
lency. 




Nitrates 


Sul- 
phates 


Chlo- 
rides 


Cya- 
nides. 


Iron 


II 


Formula. 

Chemical Equivalents 

Combining Weights 

Thermo-Chemical Equiv'lts 


Fe (N03)2 

90 
18.9 


Ve SO* 
186 
68 
12.5 


FeC18 
127 
63.5 
50 


FeCv 2 

66 
8.2 


Zinc 


II 


Formula 

Chemical Equivalents 

Combining Weights 

Thermo-Chemical Equiv'lts 


Zn(N03)» 
189 
94.5 
9.8 


ZnSO* 

161 
80.5 
11.7 


ZnCia 
186 
68 
56.4 


Zn Cy 3 
117 
58.5 
7.8 


Copper 


II 


Formula. 

Chemical Equivalents 

Combining Weights 

Thermo-Chemical Equiv'lts 


Ou(N03)2 

98.7 
7.5 


CuSO* 

159.4 

79.7 

9.3 


CuC12 

184.4 

67.2 

81.8 


CuCyS 
125.4 
62.7 
7.8 


Mercury 


II 


Formula. 

Chemical Equivalents 

Combining Weights 

Thermo-Chemical Equiv'lts 


Hg(N03)8 
824 
162 
7.5 


HgSO* 
280 
140 
9.3 


HgClS 
271 
185.5 
9.45 


HgCy2 
252 
126 
15.5 



148 
X.— CHEMICAL AND ELECTRO-CHEMICAL EQUIVALENTS. 



Name 



Hydrogen 

Gold 

Silver 

Copper (Cupric) 

Mercury (Mercuric)... 
" (Mercurous) . . 

Iron (ferric) 

11 (ferrous) 

Nickel 

Zinc 

Lead 

Oxygen 

Chlorine 



Symbols 



H 

Au 

Ag 

Cu„ 

Hg l( 

Hg, 

Fe,„ 

Fe„ 

Ni 

Zn 

Pb 



CI 



Valen 
cies 



I 
III 

I 

II 
II 

I 

III 
II 
II 
II 
II 
II 

I 



Chemical 
Equivalents 



1 

196.6 
108 

63 
200 
200 

56 

56 

59 

65 
207 

16 

85.5 



Combining 

"Weights 



1 
65.5 

108 

81.5 
100 
200 

18.7 

28 

29.5 

82.5 

103.5 

8 

85.5 



Electro- 
Chemical 
Equivalents 



.0105 
.6877 

1.134 
.8807 

1.05 

2.10 
.1964 
.294 
.3098 
.8413 

1.0868 
.084 
.8728 



XL— MAGNETIZATION AND MAGNETIC TRACTION. 



B 


B, 


Dynes 


Grammes 


Kilogrs. 


Pounds 


Lines per 


Lines per 


per 


per 


per 


per 


sq. cm. 


sq. in. 


sq. centim. 


sq. centim. 


sq. centim. 


sq. inch. 


1,000 


6,450 


89,790 


40.56 


.0456 


.577 


2,000 


12,900 


159,200 


162.8 


.1628 


2.808 


3,000 


19,850 


858,100 


865.1 


.8651 


6.190 


4,000 


25,800 


636,600 


648.9 


.6489 


9.228 


5,000 


82,250 


994,700 


1,014 


1.014 


14.89 


6,000 


88,700 


1,482,000 


1,460 


1.460 


20.75 


7,000 


45,150 


1,950,000 


1,987 


1.987 


28.26 


8,000 


61,600 


2,547,000 


2,596 


2.596 


86.95 


9,000 


58,050 


8,228,000 


8,286 


8.286 


46.72 


10,000 


64,500 


8,979,000 


4,056 


4.056 


57.68 


11,000 


70,950 


4,815,000 


4,907 


4.907 


69. 7T 


12,000 


77,400 


6,780,000 


5,841 


5.841 


83.07 


18,000 


88,850 


6,725,000 


6,855 


6.855 


97.47 


14,000 


90,300 


7,800,000 


7,550 


7.550 


113.1 


15,000 


96,750 


8,953,000 


9,124 


9.124 


129.7 


16,000 


103,200 


10,170,000 


10,890 


10.89 


147.7 


17,000 


109,650 


11,500,000 


11,720 


11.72 


166.6 


16,000 


116,100 


12,890,000 


13,140 


18.14 


186.8 


19,000 


122,550 


14,('>:X>,000 


14,680 


14.68 


208.1 


20,000 


129,000 


15,9'20,OU0 


16,280 


16.23 


280.8 



149 
Jill.— PERMEABILITY OF WROUGHT AND CAST IRON. 

BQUARB CENTIMETER MEASUREMENT. 



Annealed Wrought 


Ton. 


Gray Cast Iron. 


B 


/* 


H 


B 


M 


H 


5,000 


8,000 


1.66 


4,000 


800 


5 


9,000 


2,250 


4 


5,000 


500 


10 


10,000 


2,000 


5 


6,000 


279 


21.5 


11,000 


1,692 


6.5 


7,000 


183 


42 


12,000 


1,412 


8.5 


8,000 


100 


80 


13,000 


1,083 


12 


9,000 


71 


127 


14,000 


823 


17 


10,000 


53 


189 


15,000 


526 


28.5 


11,000 


87 


292 


16,000 


820 

161 

90 

54 


50 
105 
200 
850 








17,000 








18,000 








19,000 








20,000 


80 


666 

















SQUARE INCH MEASUREMENT. 



Annealed Wrought Iron. 


Gray Cast Iron. 


B. 


//. 


H. 


B< 


/*. 


H. 


80,000 
40,000 
50,000 
60,000 
70,000 
80,000 
90,000 


4,650 

8,877 

8,031 

2,159 

1,921 

1,409 

907 

408 

166 

76 

85 

27 


6.5 
10.3 
16.5 
27.8 
86.4 
56.8 
99.2 
245 
664 
1,581 
8,714 
5,185 


25,000 
80,000 
40,000 
50,000 
60,000 
70,000 


763 
756 
258 
114 
74 
40 


82.7 
89.7 

155 

489 

807 
1,480 


100,000 








110,000 








120,000 








180,000 








140,000 

















150 
PERMEABILITY OF SOFT CHARCOAL WROUGHT IRON 

(SHELFOBD BIDWELL.) 
SQUABE CENTIMETEB MEA8UBE. 



B 


M 


H 


7,890 


1899.1 


8.9 


11,550 


1121.4 


10.3 


15,460 


886.4 


40 


17,880 


150.7 


115 


18,470 


88.8 


208 


19,880 


45.3 


427 


19,820 


88.9 


585 



BQUAEE INCH MEAStTBEMENT. 



B, 


//. 


H. 


47,414 


1897 


25.0 


74,104 


1122 


66.1 


99,191 


888 


256 


111,189 


150 


738 


118,504 


88.8 


1385 


124,021 


45.8 


2740 


127,165 


88.9 


8753 



Both In lines of force. 



B = Magnetic Flux, 

H — Magnetizing Force, 

p 
U — 7j the Permeability or multiplying power of the core. 



.} 



151 

XIII.-MAGNETIO RELUCTANCE OF AIR BETWEEN TWO PARALLEL 

CYLINDERS OF IRON. 



b 




*• \ 




p 

Ratio of least 


CxsmarrMR Units. 


Inch Unit*. 


distance apart 








to circumference. 








0.1 


.1954 


5.1055 


0.0771 


12.968 


0.2 


.2707 


8.6917 


0.1066 


9.87T 


0.8 


.8251 


8.0768 


0.1280 


7.815 


0.4 


.8683 


2.7168 


0.1450 


6.897 


0.5 


.4046 


2.4716 


0.1593 


6.278 


0.6 


.4861 


2.2988 


0.1717 


5.825 


0.8 


.4887 


2.0465 


0.1924 


5.198 


1.0 


.5816 


1.8807 


0.2093 


4.777 


1.2 


.5684 


1.7996 


0.2238 


4.571 


1.4 


.6007 


1.6645 


0.2865 


4.228 


l.t 


.6289 


1.5902 


0.2476 


4.089 


1.8 


.6541 


1.5287 


0.2575 


8.888 


2.0 


.6774 


1.4764 


0.2667 


8.750 


4.0 


.8857 


1.1968 


0.8290 


8.040 


6.0 


.9819 


1.0782 


0.8669 


2.726 


8.0 


1.0047 


.9958 


0.3955 


8.528 


10.0 


1.0544 


.9484 


0.4151 


2.409 



In this table in columns 2 and 8 the Unit length of a cylinder is taken as 1 centi- 
meter ; in columns 4 and 6 as 1 inch, p — circumference of cylinder b — shortest 
distance apart 

XIY.-TABLE OF 6TH ROOTS. 



Num- 
ber 


Sixth 
Root 


Number 


Sixth 
Root 


Num- 
ber 


Sixth 
Root 


If umber 


Sixth 
Root 


i 


.69855 


1 


.95320 


H 


1.0177 


If 


1.0978 


1 


.70717 


1 


.96350 


i* 


1.0192 


H 


1.1019 


* 


.72306 


1 


.97006 


n 


1.0226 


IS 


1.1063 


k 


.74185 


• 


.97463 


n 


1.0260 


If 


1.1087 


i 


.76478 


1 


.97798 


H 


1.0308 


n 


1.110T 


1 


.79370 


I 


.98055 


U 


1.0379 


if 


1.1119 


I 


.88268 


A 


.98258 


li 


1.0491 


1A 


1.1129 


* 


.89090 






H 


1.0199 


2 


1.1237 


§ 


.93462 






n 


1.08S8 







152 

XV.-STANDARD AND BIRMINGHAM WIRE GAUGES. 



Standard. 


Birmingham. 


Number of 
Gauge. 


Diameter 
in Mils. 


Square of 
Diameter or 
CircTr Mils. 


Number of 
Gauge. 


Diameter 
in Mils. 


Square of 
Diameter or 
CircTr Mils, 


0000000 


500 


250000 


0000 


454 


206116 


000000 


4&4 


215296 


000 


425 


180625 


00000 


432 


186824 


00 


880 


144400 


0000 


400 


160000 





340 


115600 


000 


872 


138384 


1 


800 


90000 


00 


848 


121104 


2 


284 


80656 





824 


104976 


8 


259 


67081 


1 


300 


90000 


4 


238 


56644 


2 


276 


76176 


5 


220 


48400 


8 


252 


63504 


6 


203 


41209 


4 


232 


53S24 


7 


180 


82400 


5 


212 


44944 


8 


165 


27225 


6 


192 


86864 


9 


148 


21904 


7 


176 


80976 


10 


184 


17956 


8 


160 


25600 


11 


120 


14400 


9 


144 


20736 


12 


109 


11881 


10 


128 


16384 


13 


095 


9025 


11 


116 


13456 


14 


083 


6889 


12 


104 


10816 


15 


072 


5184 


13 


092 


8464 


16 


065 


4225 


14 


080 


6400 


17 


058 


8364 


15 


072 


5184 


18 


049 


2401 


16 


064 


4096 


19 


042 


1764 


17 


056 


8136 


20 


085 


1225 


18 


048 


2304 


21 


032 


1024 


19 


040 


1600 


22 


028 


784 


20 


036 


1296 


23 


025 


625 


21 


032 


1024 


24 


022 


484 


22 


028 


784 


25 


020 


400 


28 


024 


676 


26 


018 


824 


24 


022 


484 








25 


020 


400 








26 


018 


824 









153 



o 




a 




NH 




p 




to 




o 




(X, 




OB 

w 
O 


H 

> 


u 


H 


CO 

m 


b 

nn 


£ 


&3 


p 


fc 


3 




fa 


N 


K 


fc5 


ti 


P 


w 


g 


o 


«l 







p 


3 


H 




«4 





^ 


r 


B 


o 

QO 


m 


T-l 


M 


o 


o 




w 


& 


3 


H 


«l 


^ 


pq 


CS 


N 




o 


£ 


n 


W 


o 


fc 


< 


M 


% 


H 
(19 


u 






X 


H 




g 


< 




o 


H 


H 


fa 




U 




m 




> 




H 





0) 

ft 



as . 



<5 



• 0} Jh 






<5o 



C3 fl 



O 

3 



m o i~- \o *o 

•— i O 00 t- «o 



t-« go *?5 co .— iocot-«o«5 



$ 2 






8 S 



iO Q O iQ 
O 25 ^*< CO 



Sua os 
?o 00. o 



S3 3 



^ co 53 o t- 

co ^ o 25 i-t 

(O b> H CQ OO^ 

JO "^ CO c§ w 



g ,_ic*co^»ocot*» 



SO Q «3 U3 XO 
tO O "*• OS "£ 

CO fc~ t- co »o »o 



S 3 



8 Si S S 8 g 8 

l- «© ?o «o »« "* "* 



t- co os S> 55 

>o w -* ^ -* 



iO o o 

8 C$ 8 



8 8 



§ 8 



bo 

a 



bo 

H 

>» 



o 

8. 

GO 

d 



154 

XVII— WATTS AND HORSE POWER TABLES FOR VARIOUS 
PRESSURES AND CURRENTS. 

These tables will be found very convenient for quickly finding the 
watts and electrical horse power on lighting" and power circuits. 

To find the watts or h. p. for any current up to 1,000 amperes at a 
standard voltage add the watts or h. p. corresponding to the units, 
tens and hundreds digits of the current. 

Example : Find the electrical h. p. of 436 amperes at 105 volts. 

Solution: The h. p. for 400 amp. is 56.3, for 30 amp. it is 4.22 and for 
6 amp., .845. Adding these quantities gives 61.365 h. p., which we will 
call 61.4 h. p., as the tabular values are computed to three figures 
only, which are sufficient for engineering purposes. 

To find values for voltages higher or lower than in the tables, 
select a voltage 1-10 or ten times that required, and multiply the re- 
sult by 10 or 1-10. Thus: to find h. p. at 7 amp., 65 volts, take T amp. 
at 550 volts~5.16 h. p.; multiply by 1-10, which gives .516 h. p. To 
find h. p. at 9 amp., 1,200 volts, take 9 amp. at 120 volts=1.45 ; multi- 
ply by 10=14 5 h. p. 

To read in kilowatts place a decimal point before the watts when 
less than 1,000 in value, or substitute it for the comma in the larger 
values, 

HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 





100 volts. 


105 volts. 


110 volts. 


Amperes. 


Watts. 


h -A 


Watts. 


h.p. 


Watts. 


h.p. 


1 


100 


105 


.141 


110 


.147 


2 


200 


.268 


210 


.282 


220 


.295 


3 


300 


.402 


315 


.422 I 


330 


.442 


4 


400 


.*36 


420 


.563 


440 


.690 


5 


500 


.670 


525 


.704 


550 


.737 


6 


600 


.804 


630 


.845 


660 


.8-5 


7 


700 


.938 


• 735 


.985 


770 


1.03 


8 


800 


1.07 


840 


1.13 


880 


1.18 


9 


900 


1.21 


945 


1.27 


990 


1.33 


10 


1,000 


1.34 


1,050 


1.41 


1,100 


1.47 


20 


2,000 


2.68 


2,100 


2.82 


2,2u0 


2.95 


30 


3,000 


4.02 


3,150 


4.22 


3,300 


4.42 


40 


4,000 


5.36 


4,200 


5 63 


4,400 


6.90 


60 


5.000 


6.70 


5,250 


7.04 


5,500 


7.37 


60 


6,000 


8.04 


6,300 


8.45 


6,6('0 


8.85 


70 


7,000 


9.38 


7,350 


9.85 


7,700 


10.3 


80 


8,000 


10.7 


8.400 


11.3 


8,800 


11.8 


90 


9,000 


12.1 


9,450 


12.7 


9,9>0 


13.3 


10) 


10,000 


13.4 


10,500 


14.1 


11,000 


14.7 


200 


20,000 


26.8 


21,000 
31,500 


28.2 


22,000 


29.5 


300 


30,000 


40.2 


42.2 


33,000 


44.2 


400 


40,000 


53.6 


42,000 


56.3 


44,000 


59.0 


500 


50,000 


67.0 


52,500 


704 


55,000 


73.7 


600 


60,000 
70,000 


80.4 


03,000 


84.5 


66,000 


88.5 


700 


93.8 


73,500 


98.5 


77,000 


103 


800 


80,000 


107 


84.000 


113 


88,000 


118 


900 


90.000 


121 


94,500 


127 


99,000 


133 


1,000 


100,000 


134 


105,000 


141 


110,000 


147 



155 



HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 

{Continued.) 





115 volts. 


120 \ 


r olts. 


125 volts. 


Amp. 


Watts. 


.1^4 


Watts. 


h \?6! 


Watts. 


h.p. 


1 


115 


120 


125 


.168 


2 


230 


.308 


240 


.322 


250 


.335 


3 


345 


.462 


360 


.483 


375 


.505 


4 


460 


.617 


480 


.644 


5*0 


.670 


5 


575 


.770 


600 


.805 


625 


.840 


6 


690 


.925 


720 


.966 


750 


1.01 


7 


805 


1.08 


840 


1.13 


875 


118 


8 


920 


1.23 


960 


1.29 


1,000 


1.34 


9 


1,035 


139 


1,080 


145 


1,125 


1.51 


10 


1,150 


1.54 


1,200 


161 


1,250 


168 


20 


2.3U0 


3.08 


2,400 


3.22 


2.500 


3.35 


30 


3,450 


4.62 


3,600 


4.83 


3,750 


5.(5 


40 


4,600 


6.17 


4,800 


6.44 


5,r00 


6 70 


50 


5l750 


7.70 


6,000 


8.04 


6.25" 


8.40 


60 


6,900 


9.25 


7,200 


9 66 


7.5 


101 


70 


8.050 


10.8 


8,400 


11.3 


8,750 


11.8 


80 


9,200 


12.3 


9,600 


129 


10 000 


13.4 


90 


10 350 


13.9 


1",800 


14 5 


11,250 


15.1 


100 


11,500 


15 4 


12,00S 


16.1 


12,500 


16.8 


200 


23/00 


30.8 


24/00 


82.2 


25,000 


335 


300 


34,600 


46 2 


36,080 


48 3 


37,500 


50 5 


400 


46,000 


61.7 


48,000 


64.4 


50. f00 


67.0 


500 


57,500 


77.0 


60fK* 


80.4 


62,500 


84 


600 


69,000 


92.5 


72,000 


96.6 


75,000 


111 


700 


80 500 


1(*8 


84,000 


113 


87,500 


118 


800 


92,000 


123 


96,009 


129 


100,000 


134 


900 


103, .00 


139 


108,000 


145 


112,500 


151 


1,000 


115,000 


154 


120,000 


161 


125,*X) 


168 



15G 



HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 

(Continued.) 



Amp. 

2 

3 
4 
5 
6 

7 

8 

9 

10 

20 

30 

40 

50 

60 

70 

80 

90 

1«'0 

200 

300 

400 

500 

600 

700 

800 

9» 

1,0 



200 volts. 


Watts. 


h.p. 


200 


.268 


400 


.536 


600 


.804 


800 


1.07 


1,000 


1.34 


1,200 


1.61 


1,400 


1.88 


1,600 


2.14 


1,800 


2 41 


2,000 


2.68 


4,000 


5.36 


6,1.00 


8.04 


8.000 


10.7 


1<»,000 


13.4 


12,010 


16.1 


14,0(0 


18.8 


16,000 


21.4 


18,000 


24 1 


20,000 


26.8 


40,000 


53.6 


60,000 


80.4 


80,000 


107 


100,0(0 


134 


120,000 


161 


140,00) 


188 


160,000 


214 


180.000 


241 


200,000 


268 



210 volts. 



Watts. 

210 

420 

630 

840 

1,050 

1,260 

1,470 

1.680 

1,890 

2,100 

4,200 

6,300 

8,400 

10,500 

12,H00 

14,700 

16.800 

18,900 

21,000 

42,000 

63,000 

84,000 

105,000 

126,000 

147,000 

168,000 

189,000 

210,000 



h.p. 

.282 
.563 
.8(5 
113 
1.41 
1.69 
1.97 
2.25 
253 
2 82 
5.H3 
8 45 
11.3 
14.1 
16.9 
19.7 
22.5 
25.3 
28.2 
56.3 
84.5 
113 
141 
169 
197 
225 
253 
282 



220 \ 


olts. 


Watts. 


h.p. 


220 


.295 


440 


.690 


660 


.885 


88' I 


1.18 


1,100 


1.47 


1,320 


1.77 


1,540 


2.06 


1.760 


2.36 


1,980 


2.65 


2,200 


2.95 


4,400 


5.90 


6,600 


8.85 


8,800 


11.8 


11,000 


14.7 


13,200 


17 7 


15,400 


20.6 


17,600 


23.6 


19,800 


26.5 


22.000 


295 


44,000 


59.0 


66.000 


88.5 


88,000 


118 


110,000 


147 


132,000 


177 


154,000 


206 


176,000 


236 


198,000 


265 


220,000 


295 



is?; 



HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 

(Continued.) 





230 volts. 


240 volts. 


250 volts. 


Amp. 


Watts. 


h.p. 


Watts. 


.322 


Watts. 


.J35 


230 


.308 


240 


250 


2 


460 


.617 


480 


.644 


500 


.670 


3 


690 


.925 


720 


,966 


750 


1.01 


4 


920 


1.23 


960 


1.29 


1,000 


1.34 


5 


1,150 


1.54 


1,200 


1.61 


1,250 


1.68 


6 


1,380 


1.85 


1,440 


1.93 


1,500 


2.01 


7 


1,610 


2.16 


1,680 


2.25 


1,750 


2.35 


8 


1,840 


2.47 


1,920 


2.58 


2,000 


2.68 


9 


2,070 


2.77 


2,160 


2.90 


2,250 


3.02 


10 


2,300 


3.08 


2,400 


3.22 


2,500 


3 35 


20 


4,600 


6.17 


4,800 


6.44 


5,000 


6.70 


30 


6,900 


9.25 


7,200 


9.66 


7,500 


10.1 


40 


9,200 


12.3 


9,600 


12.9 


10,000 


13.4 


50 


11,500 


15.4 


12,000 


16.1 


12,500 


168 


60 


13,800 


18.5 


14.400 


19.3 


15,000 


20.1 


70 


16,100 


21.6 


16,800 


22.5 


17,500 


23.5 


80 


18,400 


24.7 


19,200 


25,8 


20,000 


26.8 


90 


20,700 


27.7 


21,600 


29.0 


22,500 


30.2 


100 


23,000 


30.8 


24,000 


32.2 


25,000 


33.fi 


300 


46,000 


61.7 


48.000 


64.4 


50,000 


67.0 


300 


69,000 


92.5 


72,000 


96.6 


75,000 


101 


400 


92,000 


123 


96,000 


129 


100,000 


134 


500 


115,000 


154 


120,000 


161 


125,000 


168 


600 


138,000 


185 


144,000 


193 


150,000 


201 


700 


161,000 


216 


168,000 


225 


175,0(0 


235 


800 


184,000 


247 


192,000 


258 


200,000 


268 


900 


207,000 


277 


216,000 


290 


225,000 


302 


1,000 


230,000 


308 


240,000 


322 


250,000 


335 



158 



HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 

{Continued.) 





500 volts. 


550 volts. 


600 volts. 




Watts 




Watts 




Watts 




Amp. 


and k. w. 


h Si 


and k. w. 


h.p. 


and k. w. 


h.p. 


1 


600 


550 


.74 


coo 


.804 


2 


1,000 


1.34 


1,100 


1.47 


1,200 


1.61 


3 


1,500 


2.01 


1,650 


2.21 


1,800 


2.41 


4 


2,000 


2.68 


2,200 


2.95 


2,400 


322 


5 


2,500 


3.35 


2,750 


3.69 


3,000 


4.02 


6 


3,000 


4.02 


3,300 


4.42 


3,600 


4.82 


7 


3,500 


4.69 


3,850 


5.16 


4,200 


6.63 


8 


4,000 


5.36 


4,400 


5.90 


4.800 


643 


9 


4,500 


6.03 


4,950 


6.64 


5,400 


7.24 


10 


5k.w. 


6.7 


5.5 k.w. 


7.4 


6 k.w. 


8.04 


20 


10 


13.4 


11.0 


14.7 


12 


16.1 


30 


15 


20.1 


16.5 


22.1 


18 


24.1 


40 


20 


26.8 


22.0 


29.5 


24 


32.2 


50 


25 


33.5 


27.5 


36.9 


30 


40.2 


60 


30 


40.2 


33.0 


44.2 


36 


48.2 


70 


35 


46.9 


38.5 


51.6 


42 


66.3 


80 


40 


636 


44.0 


59.0 


48 


64.3 


90 


45 


60.3 


49.5 


66.4 


54 


72.4 


100 


50 


67 


55 


74.0 


60 


80.4 


200 


100 


134 


110 


147 


120 


161 


300 


150 


201 


165 


221 


180 


241 


400 


200 


268 


2B0 


295 


240 


322 


500 


250 


336 


275 


369 


300 


402 


600 


300 


402 


330 


442 


360 


483 


700 


350 


469 


885 


616 


420 


563 


800 


400 


536 


440 


590 


480 


643 


900 


450 


603 


495 


664 


540 


724 


1,000 


500 


670 


550 


740 


600 


804 



INDEX. 



Alternating current sys- 
tem 47-49 

Alternating currents .... 115 
Amperage of armature... 99 

Ampere 11-12 

Ampere-second 57 

Ampere-turns 87 

Ampere-turns, rules for 

calculating 87 

Armature amperage of, on 

short circuit 99 

Armature calculations. .99-103 
Armature, capacity of . . . . 99 
Armature, general fea- 
tures of 96 

Armature, resistance of . . 98 
Armatures, rules for cal- 
culating .98-99 

Armature, voltage of . . . . 99 
Armature winding iron or 
copper 96 

Bridge, Wheatstone, prin- 
ciple of 25 

Batteries, illustrations of 
arrangements 66 

Batteries or generators in 
opposition 15 

Batteries, storage, resist- 
ance of 65 

Battery arrangement and 
size for given efficiency. 73 

Battery, arrangement of 
cells in 67 

Battery calculations, dis- 
crepancies in 72 

Battery, chemicals con- 
sumed in 78 

Battery constants 65 

Battery, current of 68 

Battery, effective rate of 
work of 77-78 



Battery, efficiency, to cal- 
culate 72 

Battery, electromotive 

force of 67 

Battery, how rated 65 

Battery, resistance of . . . . 67 
Battery, to calculate cur- 
rent and arrangement. 69-72 
Battery, to calculate its 

voltage 76-77 

Battery, work of 77-78 

Calorie, gram and kilo- 
gram, defined 54 

Calorie, relation to watts 

and ergs 54-55 

Capacity of armature 

winding 97 

Cars, power to move .... Ill 

Cell constants 65 

Chemicals consumed in a 

battery 78 

Circuits, divided, branched 

or shunt 19-25 

Circuits, portions of . . . .17-19 
Circuits, single conductor, 

closed 14-15 

Circular mils calcula- 
tions 44-45 

Circular mils 43-45 

Condensers 121 

Conductance 36-37 

Conductors of same mate- 
rial 26 

Conductors, resistance of 

different 26 

Contact, area of in elec- 
tro-magnets and arma- 
tures 86 

Convection and radiation, 

loss of heat by 58 

Conversion, ratio of 47 



160 



INDEX. 



Converters, rule for wind- 
ing 40 

Cores of field-magnet, to 
calculate 104 

Counter-electromotive force 
of plating-bath 80 

Current, battery required 
for a given 60-72 

Current, distribution in 
parallel circuits 20-21 

Current, its heating effect 
and energy 50 

Currents passed by 
branches of a circuit.. 10 

Current, work of 60 

Current yielded by a bat- 
tery 65 

Definitions, general 9 

Demonstrations of rules. 127 

Derived units 10 

Dimensions, units of 10 

Drop of potential in 

leads 38-41 

Drum type closed circuit 

armatures 08-99 

Duty and commercial ef- 
ficiency 63-64 

Duty of generators 63 

Effective E. M. F 16 

Eniciency, its relation to 
resistances 64 

Eniciency of generators, 
commercial 64 

Efficiency of generators, 
electrical 63 

Efficiency, to calculate 
battery for a given. . 73-74 

Electrical railways 109 

Electro-magnets and dy- 
namos 82 

Electro-magnets, general 
rules for 85 

Electro-magnets, to mag- 
netize 86-87 

Electro-magnets, traction 
of 85-86 

Electromotive force of 
battery 67 

Electro - plating calcula- 
tions 79 

Energy defined 9 

Energy in circuit, rules 
for determining 50-53 

Energy of current 50-53 



E. M. F,* its meaning. . . 10 
E. M. F., primary and sec- 
ondary 47 

Erg 54 

Force defined 9 

Force, magnetic 84 

Fundamental units 10 

Field-magnet cores, to cal- 
culate 104 

Field-magnet for dynamo 

or motor 104-107 

Field, unit intensity of 
magnetic 82-83 

Generators, eniciency of, 

63-64 
Generators or batteries in 
opposition 15 

Heat, absolute quantity in 

circuit 54 

Heating effect of current, 

50-53 

Heat, specific 57 

Horse-power, electrical . . 61 
Horse - power, reduction 

from kilogram-meters. . 60 
Horse-power, to calculate 
for lamps 62 

Joule, his law of heating, 

effect of currents 50 

Joule or gram-calorie. ... 54 

Kapp, Gisbert, his line of 

force 107-108 

Kilogram-meters 60 

Leads, tapering in size . . 41-42 
Leakage between cylindri- 
cal magnet leg, to cal- 
culate 91 

Leakage between flat mag- 
net surfaces, to calcu- 
late 91 

Leakage of lines in a mag- 
netic circuit 95 

Leakage of lines of force. 89 
Line of force, Kapp's.. 107-108 

Lines of force 82 

Lines of force cut per sec- 
ond for one volt 96 

Lines of force, leakage of. 89 
Lines of force, to diminish 
leakage of 94-95 



INDEX. 



161 



Magnetic circuit 84 

Magnetic circuit calcula- 
tions 88-80 

Magnetic circuit, the law 

of 85 

Magnetic circuits, four 

parts of 88 

Magnetic circuits, general 

calculations for 92 

Magnetic field 82-83 

Magnetic flux 82-83 

Magnetic force 84 

Magnetic potential, aver- 
age difference of 90 

Magnetism, no insulator of 84 

Magnetizing force 87 

Magnet legs, long and 

short 94 

Mass defined 9 

Metals, deposition of, by 

battery 79 

Mho, unit of conductance 36 
Mil, circular, calculations 

based on 44-45 

Mils, circular 43 

Mils, circular, applied to 

alternating current ... 48 
Multiple arc connections, 
to calculate 38-42 

Neuteal wire in three 

wire system 46 

Notation in powers of ten 136 

Ohm 11-12 

Ohm's law 13-25 

Ohm's law, its universal 

application 14 

Ohm's law, six expres- 
sions of 13-14 

Pakallel connections of 

equal resistance 23 

Parallel leads, resistance 

of 22 

Permeance 83-84 

Permeability 85 

Permeability, average 

range of 104 

Potential, diagram for cal- 
culating fall of 41 

Potential difference 38-42 

Potential difference, drop, 

or fall of 17 

Potential, drop of in leads, 

38-42 



Power to move cars Ill 

Powers of ten, notation in 130 

Primary LL M. F 47 

Proving armature calcula- 
tions 103 

Radiation and convection, 

loss of heat by 58 

Railways, electric 109 

Rate of heat-energy, units 

of 63 

Ratio of conversion 47 

Reluctance 83-84 

Reluctance, calculation of. 87 

Resistance 26-35 

Resistance and efficiency, 

how related 64 

Resistance defined 9 

Resistance of battery.... 67 
Resistance of circuit, note 

relative thereto 14 

Resistance of parallel 

leads 22-23 

Resistance referred to 
weight of conductor. . .35-36 

Resistance specific 29 

Resistances, two in bat- 
tery circuits 65 

Resistance, universal rule 

for 31-32 

Rules, demonstrations of. 127 

Safety-catches for fuses, 

how calculated 59 

Secondary E. M. F 47 

Self-Induction 117 

Series winding for dyna- 
mos 106 

Shunt circuits 19 

Shunt circuits, resistance 

of 22 

Shunt winding for dyna- 
mos 107 

Sizes of feeders 109 

Space defined 9 

Specific heat 57 

Specific resistance 30-31 

System, alternating cur- 
rent 47-49 

Systems, special 46-49 

System, three wire 46-47 

Tables — 

American wire gauge 
table 146 



162 



INDEX. 



Tables — Con tinned. 

Chemical and electro- 
chemical equivalents . . 148 

Chemical and thermo- 
chemical equivalents . . 147 

Current capacity of bare 
or insulated overhead 
wires 153 

Equivalents of units ot 
area 140 

Equivalents of units of 
energy and work. . .142-143 

Equivalents of units of 
length 139 

Equivalents of units of 
volume 140 

Equivalents of units of 
weight 141 

Magnetic reluctance of air 
between two parallel 
cylinders of iron 151 

Magnetization and mag- 
netic traction 148 

Permeability of soft char- 
coal wrought iron .... 150 

Permeability of wrought 
and cast iron 149 

Relative resistance and 
conductance of pure 
copper at different tem- 
peratures 145 

Specific resistance of solu- 
tions and liquids 144 

Standard and Birmingham 
wire gauges 152 

Table of specific resist- 
ances in microhms and 



of coefficients of specific 

resistances of metals. . 144 

Table of the sixth roots. . 1 r> 1 

Three wire system 48 

Three wire system, saving 

in size of wires 4G 

Three wire system, the 

neutral wire 4G 

Time defined 9 

Units, concrete statement 

of 12 

Units, fundamental 10 

Units, original and de- 
rived 10 

Volt 10-12 

Voltage of battery, to cal- 
culate . . . ., 76-77 

Volt amperes 56-60 

Watt 56-60 

Watts and horse power 
tables for various pres- 
sures and currents. . . . 154 

Weight defined 9 

Weight of conductor, re- 
sistance referred to... 35-36 
Wheatstone bridge, prin- 
ciple of 25 

Winding, series and 

shunt 106-107 

Wire, rule for heating of, 

by a current 59 

Work defined 9 

Work of current 60 



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